Practice Questions: Arguments & Propositions in NL, and Computational Argumentation

Extracted from practices/*.pdf (via tmp/txt/mt_mid_term_practice_*.txt), filtered to the two topics covered in arg_and_computational_argumentation.md. Each question keeps its original wording, answer, and a worked explanation.

Note: the mt-number tags used in full_03.md’s answer key do not line up with the physical practice file suffixes (verified by grep). Argumentation-framework content actually lives in mt_mid_term_practice_4/5/7.txt (not _1.txt), and the barber-paradox question is in mt_mid_term_practice_6.txt. Sections below are tagged by the file the question was read from.

How each question is laid out:

Answer: the short result, as a quote block.
Worked solution: numbered steps, one idea per step.
A picture (argument map, table, or Venn diagram) where it helps.

Part 1: Arguments & Propositions in Natural Language

mt1 (mid term practice 1.pdf)

Q1 (4 marks): Fill in the blanks

“The statements that make up an argument are divided into one or more premises and one and only one conclusion. The premises are the statements that set forth the reasons or evidence, and the conclusion is the statement that is claimed to follow from the premises.”

Answer: premises (reasons/evidence) and conclusion (what is claimed to follow).

Worked solution

Recall the definition of an argument
- An argument is a set of statements with two roles.
- Premises: the reasons or evidence given (one or more).
- Conclusion: the one claim said to follow from them.
Match the blanks to the definition
- Blank 1 asks what the reason/evidence statements are called: premises.
- Blank 2 asks what the statement claimed to follow is called: conclusion.
Confirm
- The passage restates the definition, so the fill-in is a direct reading.

 premise 1 ┐
 premise 2 ├──▶ conclusion      (one or more premises, exactly one conclusion)
 premise n ┘

Q2 (4 marks): Mark reason (R) and conclusion (C)

For each passage below, mark the reason(s) with R(…) and the conclusion(s) with C(…).

(a) Since the good, according to Plato, is that which furthers a person’s real interests, it follows that in any given case when the good is known, men will seek it. (b) The fact that there was never a land bridge between Australia and mainland Asia is evidenced by the animal species in the two areas being very different. Asian placental mammals and Australian marsupial mammals have not been in contact in the last several million years. (c) Women of the working class should not have more than two children at most. The average working man can support no more and the average working woman can take care of no more in decent fashion. (d) Many people believe a dark tan is attractive and healthy, but mounting evidence indicates too much sun can lead to health problems. One noticeable effect is premature aging of the skin, and it plays a role in skin cancer.

Answer:

(a) Since R(the good, according to Plato, is that which furthers a person’s real interests), it follows that C(in any given case when the good is known, men will seek it).

(b) The fact that C1-R2(there was never a land bridge between Australia and mainland Asia) is evidenced by R1(the animal species in the two areas are very different). C2(Asian placental mammals and Australian marsupial mammals have not been in contact in the last several million years).

(c) C(Women of the working class should not have more than two children at most). R(The average working man can support no more and the average working woman can take care of no more in decent fashion).

(d) Many people believe a dark tan is attractive and healthy, but mounting evidence indicates C(too much sun can lead to health problems). One noticeable effect is R(premature aging of the skin… it plays a role in skin cancer).

Note: (b) is a two-step chain. R1 supports C1, and C1 then acts as R2 for C2.

Note (check): in (b) the key reads “not in contact for millions of years” as a conclusion (C2). It could also be read as the explanation of why the species differ. The key’s reading is kept.

Argument maps

(a)  R  the good = what furthers real interests        ("Since")
        ──▶  C  when the good is known, men seek it    ("it follows that")

(b)  R1 species in the two areas differ                ("is evidenced by")
        ──▶  C1/R2 no land bridge ever existed
                 ──▶  C2 no contact for millions of years

(c)  C  working-class women: at most two children      (stated first, no indicator)
        ◀──  R  a man can support, a woman can care for, no more

(d)  (opinion: dark tan is healthy)   "but" = contrast, not tagged
     R  premature skin aging, skin cancer              ("One noticeable effect is")
        ──▶  C  too much sun can cause health problems ("mounting evidence indicates")

Worked solution

Look for indicator words
- Conclusion indicators: therefore, so, hence, thus, it follows that.
- Premise indicators: since, because, for, given that, evidenced by.
Tag each passage
- (a) “since” marks the premise, “it follows that” marks the conclusion.
- (c) No indicator, but the claim comes first and the reason follows.
- (d) “mounting evidence indicates” introduces the conclusion; “one noticeable effect is” introduces its support.
Handle the chain in (b)
- “X is evidenced by Y” means Y is the evidence (R1) and X is what it supports (C1).
- X then serves as a reason (R2) for the broader claim C2.
- A statement can be the conclusion of one step and a premise of the next.
Check each tag
- Read it as “because [premise], [conclusion]“. If it sounds natural, the tags are right.

Q3 (4 marks): Extract main conclusion from editorial text

(a) [team-sports/children’s fitness passage]. (b) [elective cesarean section passage].

Answer:

(a) “Our inappropriate concept of team sports causes the health and fitness problem of our children” (or weaker: “Our concept of team sports is inappropriate”).

(b) “Elective cesarean section should be restrained.”

Worked solution

Know what “main conclusion” means
- It is the one central claim the author argues for.
- Everything else is a premise or a sub-conclusion feeding it.
Ask one test question of each statement
- Is it offered as a reason for something else, or is it what everything else tries to establish?
- A statement that is supported but supports nothing further is the main conclusion.
Accept paraphrase
- Editorials often leave the main conclusion implicit.
- Paraphrasing the overall thrust is normal for non-standard-form arguments.
Apply to (a)
- The passage criticises the current concept of team sports and links it to children’s poor health.
- The tying claim: the concept of team sports is the cause of the problem.
Apply to (b)
- The passage presents harms and societal concerns about elective cesareans.
- The tying claim: elective cesarean section should be limited.

Q4 (4 marks): Which passages are arguments; identify conclusion

For each passage, state (i) whether it is an argument, and (ii) if so, what the conclusion is.

(a) “If public education fails, then it is likely that the democratic government will fail as well.” (b) “It is strongly recommended that you have your house inspected for termites before purchasing it.” (c) “Mosquito bites are not always the harmless little irritations most people think. Some mosquitoes carry the West Nile virus, and others carry encephalitis.” (d) “Young people at universities study to achieve knowledge and not to learn a trade. Therefore, universities should teach only theoretical subjects and abandon all trade-oriented programmes.”

Answer:

(a) Not an argument (conditional sentence).

(b) Not an argument (a recommendation/directive, not an evidenced conclusion).

(c) An argument; conclusion = “Mosquito bites are not always the harmless little irritations most people think.”

(d) An argument; conclusion = “Universities should teach only theoretical subjects and abandon trade-oriented programmes.”

Verdict table

Item	Argument?	Deciding feature
(a)	no	”If P then Q” asserts neither P nor Q
(b)	no	directive; no reason offered
(c)	yes	facts about West Nile and encephalitis support the opening claim
(d)	yes	”Therefore” marks the conclusion

Worked solution

Recall the test
- A passage is an argument only if at least one premise is offered for a conclusion.
- Premise and conclusion must be genuine claims (true or false).
Test (a)
- A conditional “If P then Q” does not assert P or Q.
- It only asserts a relationship, so no premise is given.
Test (b)
- A recommendation is a directive, not a claim with a premise-conclusion structure.
- Reasoning may sit behind it, but none is stated.
Test (c)
- The facts about the two viruses are reasons for the opening claim.
- Ask “what are the facts leading to?” to find the conclusion.
Test (d)
- The reason (“study to achieve knowledge, not a trade”) supports a prescriptive conclusion.
- “Therefore” flags it.

Q5 (4 marks): Deductive or inductive?

Classify each argument as deductive or inductive and briefly justify your answer.

(a) Amoco is listed on the New York Stock Exchange. Exxon is listed. Texaco is listed. Therefore, all major American oil companies are listed on the New York Stock Exchange. (b) A pendulum that is longer swings more slowly than a shorter one. The pendulum of this clock has been lengthened. Therefore, this clock will run more slowly. (c) The word “cholesterol” is derived from the Greek “chole” (bile) and “stereos” (solid). Cholesterol is endogenous to the human body. Therefore, cholesterol is manufactured inside the human body. (d) If Alexander the Great died of typhoid fever, he must have been infected while in India. Alexander the Great did die of typhoid fever. Therefore, he was infected while in India.

Answer:

(a) Inductive.

(b) Deductive.

(c) Inductive (no penalty if answered Deductive; this is the explicitly flagged ambiguous case).

(d) Deductive.

Verdict table

Item	Verdict	Deciding feature
(a)	inductive	generalises from three examples to “all”
(b)	deductive	general rule applied to a specific case
(c)	inductive (deductive also accepted)	close to definitional but no universal rule stated
(d)	deductive	modus ponens: If P then Q; P; so Q

 deductive: premises true  ═▶ conclusion cannot be false
 inductive: premises true  ──▶ conclusion probably true

Worked solution

Recall the distinction
- Deductive: if the premises are true, the conclusion cannot be false.
- Inductive: the premises make the conclusion likely, not certain.
Test (a)
- Three companies are listed; the conclusion says “all major companies”.
- It goes beyond the evidence, so it is a generalisation from examples: inductive.
Test (b)
- The rule “longer swings slower” plus “this pendulum was lengthened” gives the conclusion by direct instantiation.
- The conclusion follows with necessity: deductive.
Test (c)
- “Endogenous” means originating from within, so the step is nearly definitional.
- The inference is not spelled out as a universal rule, so it is marked inductive.
- Deductive is also accepted: this is the flagged ambiguous case.
Test (d)
- Standard modus ponens: (If P then Q), P, so Q.
- The conclusion follows necessarily: deductive.

Q11 (5 marks): Best categorical form (A/E/I/O) for each NL proposition

Restate each sentence in the best standard categorical form (A: “All S are P”, E: “No S are P”, I: “Some S are P”, or O: “Some S are not P”) and identify the form letter.

(a) “Every dog is a mammal.” (b) “Cats never bark.” (c) “None of the students failed.” (d) “Whoever studies hard passes the exam.” (e) “Fish don’t have legs.”

Answer:

(a) All S are P (A).

(b) No S are P (E).

(c) No S are P (E).

(d) All S are P (A).

(e) No S are P (E).

Cheat sheet: natural-language pattern to A/E/I/O

NL pattern	Form	Note
Every / all / each / any S is P	A: All S are P	universal, affirmative
Whoever / anyone who S is P	A: All S are P	a conditional in disguise
Only S are P	A: All P are S	subject and predicate swap
The only ones who P are S	A: All P are S	same swap as “only”
No S is P	E: No S are P	universal, negative
S never P / none of the S P / S don’t P (no exception)	E: No S are P	”never” and “none” are universal negatives
Some S is P / there is an S that is P / somebody P	I: Some S are P	particular, affirmative
Some S are not P / not all S are P / somebody does not P	O: Some S are not P	particular, negative

Applying the cheat sheet

Sentence	Keyword trap	S	P	Form
(a) Every dog is a mammal	”every” = all	dogs	mammals	A
(b) Cats never bark	”never” = universal negative	cats	barking things	E
(c) None of the students failed	”none of” = no	students	people who failed	E
(d) Whoever studies hard passes the exam	”whoever” = all who (conditional)	hard-studiers	exam-passers	A
(e) Fish don’t have legs	plain denial, no exception stated	fish	leg-having things	E

Worked solution

Recall the four templates
- A: All S are P. E: No S are P. I: Some S are P. O: Some S are not P.
- Every sentence must be reworded into exactly one before it can be diagrammed or tested (see Section 2 of the theory file).
Apply to (a)
- “Every dog is a mammal” already has the universal-affirmative shape.
- Standard form: All dogs are mammals (A).
Apply to (b) and (c)
- “Never” and “none” deny the predicate of every member of the subject class.
- Standard forms: No cats are barking things (E); No students are people who failed (E).
Apply to (d)
- “Whoever studies hard passes” means “if S then P” for every individual.
- That is the same as “All S are P”. Standard form: All hard-studiers are exam-passers (A).
Apply to (e)
- “Fish don’t have legs” universally denies “have legs” of fish, with no stated exception.
- Standard form: No fish are leg-having things (E).

mt2 (mid term practice 2.pdf)

Q1: Is it an argument?

State whether each passage is an argument and explain why.

1.1 “Charles went bald, and most men go bald.” 1.2 “Toward evening, clouds formed and the sky grew darker; then the storm broke.” 1.3 “The hour is up, so you must hand in your exams.”

Answer:

1.1 Not an argument.

1.2 Not an argument.

1.3 An argument.

Verdict table

Item	Argument?	Deciding feature
1.1	no	two facts joined by “and”; neither supports the other
1.2	no	narrative sequence; “then” marks time, not reason
1.3	yes	”so” is a conclusion indicator

1.3   R "The hour is up"  ──▶  C "you must hand in your exams"      ("so")

Worked solution

Identify the test
- It is an argument when one statement is offered as a reason for another.
- A listing or narration of facts, even with “and” or “then”, is not an argument.
Test 1.1
- “Charles went bald” and “most men go bald” are simply conjoined.
- Neither is presented as a reason for the other.
Test 1.2
- “Then” describes what happened next, not a logical or evidential link.
- No conclusion is drawn: it is a narrative.
Test 1.3
- “The hour is up” is the reason; “you must hand in your exams” is the conclusion.
- “So” explicitly marks the inference: an argument.

Q2: Extract main conclusion from editorial

Identify the main conclusion of each passage.

2.1 “When the detective arrived at the scene of the burglary, she noticed pry marks on the rear door and muddy footprints near the back threshold. Both the rear door and the front door were found unlocked and slightly open.” 2.2 “Stephen Hawking has argued that the second law of thermodynamics (which says entropy always increases) entails that the arrow of time points from the past (low entropy) to the future (high entropy). Since the universe is expanding and entropy is increasing, the universe must have been in a more ordered state at its beginning.” 2.3 “When two linear equations are graphed, their solution is the point where the lines intersect. However, because the lines are plotted by hand on graph paper, the exact intersection can only be estimated visually. The finer the scale, the better the estimate, but it is always an estimate.”

Answer:

2.1 “The thief entered through the rear door and left through the front.”

2.2 “The condition of the universe at the instant of the Big Bang was more highly ordered than it is today.”

2.3 “The graphical method for solving a system of equations is an approximation.”

2.1  pry marks, muddy footprints, both doors open ──▶ C: thief in by the rear door, out by the front
2.2  entropy rises, universe expands              ──▶ C: the Big Bang state was more highly ordered
2.3  hand-plotted lines, estimated intersection   ──▶ C: the graphical method is an approximation

Worked solution

Use the same technique as mt1 Q3
- Find the claim the rest of the passage is evidence for.
- It may sit mid-passage, after phrases like “it must be the case that”, “this shows that”, “we can conclude”.
Apply to 2.1
- The pry marks, footprints and open doors are premises.
- Together they point to one reconstruction: the thief’s path through the house.
Apply to 2.2
- The passage builds from entropy and the arrow of time.
- The claim it is designed to establish is the ordered state of the universe at the Big Bang.
Apply to 2.3
- The remarks on graphical intersections and estimation error all support one claim.
- That claim is that the method gives an approximation, not an exact solution.

Q3: Syllogism validity by Venn diagram

“All squares have equal sides. Some squares are rectangles. ∴ Some rectangles have equal sides.”

Answer: Deductive (valid).

Region key (used by every syllogism question below)

 Three circles S, P, M drawn as overlapping boxes, 7 regions:
   1 = S only          2 = S and P only        3 = P only
   4 = S and M only    5 = P and M only        6 = S and P and M       7 = M only
 #   = empty (shaded by a universal premise)
 *   = at least one thing (from a particular premise)
 *   on a line = could be in either neighbouring region (undetermined)
 Order of drawing: universal premise first, then the particular one.

Terms: M = squares (middle), S = rectangles (subject of conclusion), P = equal-sided things (predicate).

0. blank                    1. P1: All M is P           2. P2: Some S is M
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
| S      |     |      P |   | S      |     |      P |   | S      |     |      P |
|   1    |  2  |   3    |   |   1    |  2  |   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+-----+---+    |
|   | 4  |  6  | 5 |    |   |   |#4##|  6  | 5 |    |   |   |#4##|  6  | 5 |    |
|   |    |     |   |    |   |   |####|     |   |    |   |   |####|  *  |   |    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |######7#######|            |######7#######|
    | M            |            |#M############|            |#M############|
    +--------------+            +--------------+            +--------------+

 P1 All M is P   : M outside P is empty        -> regions 4, 7 shaded
 P2 Some S is M  : star in S and M = 4 or 6    -> 4 is empty, so star in 6
 Conclusion: Some S is P needs a star in 2 or 6.   Found: star in 6.

Verdict: valid

Worked solution

Assign the three terms
- M appears in both premises but not in the conclusion.
- Draw the 7-region layout from Section 2 of the theory file.
Diagram premise 1 first (universal)
- “All M is P” empties every part of M outside P: regions 4 and 7.
Diagram premise 2
- “Some S is M” needs a star in S and M, that is region 4 or 6.
- Region 4 is already empty, so the star is forced into region 6.
Read off the conclusion
- “Some S is P” holds when there is a star in region 2 or 6.
- The star in 6 guarantees it: the syllogism is deductively valid.

Q4: Syllogism validity

“All pediatricians are doctors. All pediatricians like children. ∴ All doctors like children.”

Answer: Not deductive (invalid).

Terms: M = pediatricians, S = doctors, P = people who like children.

0. blank                    1. P1: All M is P           2. P2: All M is S
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
| S      |     |      P |   | S      |     |      P |   | S      |     |      P |
|   1    |  2  |   3    |   |   1    |  2  |   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+-----+---+    |
|   | 4  |  6  | 5 |    |   |   |#4##|  6  | 5 |    |   |   |#4##|  6  |#5#|    |
|   |    |     |   |    |   |   |####|     |   |    |   |   |####|     |###|    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |######7#######|            |######7#######|
    | M            |            |#M############|            |#M############|
    +--------------+            +--------------+            +--------------+

 P1 All M is P : M outside P is empty   -> 4, 7 shaded
 P2 All M is S : M outside S is empty   -> 5, 7 shaded (7 already)
 Conclusion: All S is P needs S outside P empty = regions 1 and 4.
 Region 4 is shaded, region 1 is still open.

Verdict: invalid

Worked solution

Assign terms
- M = pediatricians, S = doctors, P = child-likers.
Diagram premise 1
- “All M is P” empties M outside P: regions 4 and 7.
Diagram premise 2
- “All M is S” empties M outside S: regions 5 and 7 (7 was already empty).
Check the conclusion
- “All S is P” needs regions 1 and 4 both empty.
- Region 1 (doctors who are neither pediatricians nor child-likers) is untouched, so the conclusion is not forced.
- Intuition: a doctor who is not a pediatrician need not like children.

Q5: Syllogism validity

“No P is M. No M is S. ∴ No S is P.”

Answer: Not deductive (invalid).

Terms: here P is the predicate of the conclusion, M the middle term, S the subject. Premise 1 relates P and M; premise 2 relates M and S.

0. blank                    1. P1: No P is M            2. P2: No M is S
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
| S      |     |      P |   | S      |     |      P |   | S      |     |      P |
|   1    |  2  |   3    |   |   1    |  2  |   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+-----+---+    |
|   | 4  |  6  | 5 |    |   |   | 4  |##6##|#5#|    |   |   |#4##|##6##|#5#|    |
|   |    |     |   |    |   |   |    |#####|###|    |   |   |####|#####|###|    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |      7       |            |      7       |
    | M            |            | M            |            | M            |
    +--------------+            +--------------+            +--------------+

 P1 No P is M : overlap of P and M is empty  -> 5, 6 shaded
 P2 No M is S : overlap of M and S is empty  -> 4, 6 shaded (6 already)
 Conclusion: No S is P needs the S and P overlap empty = regions 2 and 6.
 Region 6 is shaded, region 2 is still open.

Verdict: invalid

Worked solution

Assign terms
- P is the conclusion’s predicate, M the middle term, S the subject.
Diagram premise 1
- “No P is M” empties the P and M overlap: regions 5 and 6.
Diagram premise 2
- “No M is S” empties the M and S overlap: regions 4 and 6 (6 already empty).
Check the conclusion
- “No S is P” needs the whole S and P overlap empty: regions 2 and 6.
- Region 2 (both S and P, but not M) is never touched, so the conclusion is not forced.

Q6: Syllogism validity

“No P is M. Some S is M. ∴ Some S is not P.”

Answer: Deductive (valid). (Identical wording appears again as mt3 Q4 and mt3 Q7 later in this file; all three share this one derivation.)

Terms: M = middle term, S = subject of conclusion, P = predicate of conclusion. This is Example 2.1.1 in Section 2 of the theory file.

0. blank                    1. P1: No P is M            2. P2: Some S is M
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
| S      |     |      P |   | S      |     |      P |   | S      |     |      P |
|   1    |  2  |   3    |   |   1    |  2  |   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+-----+---+    |
|   | 4  |  6  | 5 |    |   |   | 4  |##6##|#5#|    |   |   | 4  |##6##|#5#|    |
|   |    |     |   |    |   |   |    |#####|###|    |   |   | *  |#####|###|    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |      7       |            |      7       |
    | M            |            | M            |            | M            |
    +--------------+            +--------------+            +--------------+

 P1 No P is M    : P and M overlap is empty     -> 5, 6 shaded
 P2 Some S is M  : star in 4 or 6; 6 is empty   -> star in 4
 Conclusion: Some S is not P needs a star in S outside P = region 1 or 4.
 Found: star in 4.

Verdict: valid

Worked solution

Assign terms
- M middle, S subject, P predicate.
Diagram premise 1 (universal, first)
- “No P is M” empties regions 5 and 6.
Diagram premise 2 (particular)
- “Some S is M” needs a star in region 4 or 6.
- Region 6 is empty, so the star is forced into region 4.
Read off the conclusion
- “Some S is not P” holds when there is a star in region 1 or 4.
- The star in 4 guarantees it: valid.

mt3 (mid term practice 3.pdf)

Q1: Identify argument + standard form

For each passage: (i) state whether it is an argument, and (ii) if it is, write it in standard form (list premises, then draw the ”∴” line, then state the conclusion).

1.1 “Since Chicago is north of Boston, and Boston is north of Charleston, Chicago is north of Charleston.” 1.2 “Joe quit, because his boss was giving him so much grief.” 1.3 “Witches float because witches are made of wood, and wood floats.” 1.4 “Both houses of Congress may pass a bill, but the president may still veto it.”

Answer: 1.1, 1.2 and 1.3 are arguments; 1.4 is not.

Argument maps and standard forms

1.1   R1 Chicago is north of Boston      ┐ "Since"
      R2 Boston is north of Charleston   ┴──▶ C Chicago is north of Charleston

      Chicago is north of Boston
      Boston is north of Charleston
      ─────────────────────────────────
      ∴ Chicago is north of Charleston

1.2   C Joe quit  ◀── R his boss was giving him so much grief       ("because")

      His boss was giving him so much grief
      ─────────────────────────────────────
      ∴ Joe quit

1.3   C Witches float ◀── R1 witches are made of wood  ┐ ("because")
                          R2 wood floats               ┘

      Witches are made of wood
      Wood floats
      ─────────────────────────
      ∴ Witches float

1.4   "Congress may pass" , "but" , "president may veto"
      two independent facts in tension: no arrow, no argument

Worked solution

Look for indicator words
- Premise indicators: since, because, for.
- Conclusion indicators: therefore, so, thus.
Test 1.1
- “Since” introduces both premises; the conclusion follows by transitivity.
- List premises above the therefore line.
Test 1.2 and 1.3
- “Because” introduces the premise(s); the conclusion is stated first.
- Standard form reverses the order: premises on top, conclusion below.
Test 1.4
- “But” is a contrast word, not a logical indicator.
- The two clauses are independent facts: neither is evidence for the other.
- This is a conjunction, not an argument.

Q2: Suppressed premises, standard form

Each argument below has one or more unstated (suppressed) premises. Supply the missing premise(s) and rewrite the argument in standard form.

2.1 “In The Netherlands the electrical outlets are above floor level because of flooding danger.” 2.2 “Mercury is a liquid at room temperature and hence this pound of mercury would be liquid in this room.” 2.3 “Joe is not a freshman because he studies in Bangkadi campus.” 2.4 “In some situations, in order to uphold his oath, the President of the U.S. may have to take immediate action.”

Answer: see the standard forms below; lines marked [suppressed] were not in the original text.

2.1   (1) Flooding is a danger in the Netherlands.                                    [stated]
      (2) In countries where flooding is a danger, it is safer to place
          electrical outlets above floor level.                                       [suppressed]
      ─────────────────────────────────────────────────────────────────────────
      ∴ In the Netherlands the electrical outlets are above floor level.

2.2   (1) Mercury is a liquid at room temperature.                                    [stated]
      (2) This room is at room temperature.                                           [suppressed]
      ─────────────────────────────────────────
      ∴ This pound of mercury would be liquid in this room.

2.3   (1) Joe studies in Bangkadi campus.                                             [stated]
      (2) Freshmen study in Bangkadi campus.                                          [suppressed, as given in source]
      ─────────────────────────────────────────
      ∴ Joe is not a freshman.

2.4   (1) The President of the U.S. takes an oath to ensure the security of the U.S.  [suppressed]
      (2) Some situations might require immediate action to ensure the security
          of the U.S.                                                                 [suppressed]
      ─────────────────────────────────────────────────────────────────────────
      ∴ In some situations, in order to uphold his oath, the President of the U.S.
        may have to take immediate action.

Note (check): 2.3’s suppressed premise, as given in the source key, does not strictly yield “Joe is not a freshman”. The premise “Students who study in Bangkadi campus are not freshmen” would. The source wording is kept.

Worked solution

Understand enthymemes
- An enthymeme leaves one or more premises unstated because the audience shares them.
- To rebuild it, supply the “bridging” premise linking stated premise to conclusion.
Method
- Ask: what general rule, combined with the stated premise, yields the conclusion?
- That rule is the suppressed premise.
Apply to 2.1 and 2.2
- 2.1 bridge: where flooding is a danger, outlets should be above floor level.
- 2.2 bridge: this room is at room temperature; without it the general fact about mercury does not reach this room.
Apply to 2.3
- The source gives “Freshmen study in Bangkadi campus” as the suppressed premise (see the note above).
- Strict validity would need “students who study in Bangkadi campus are not freshmen”.
Apply to 2.4
- Both premises are suppressed.
- The first states the oath’s content (security of the U.S.); the second links “some situations” to a need for immediate action.
- Together they give the stated conclusion.

Q3: Extract 2+ arguments from text, using argument-map hints

For each passage: (i) identify and write out at least two distinct arguments in standard form, (ii) note any counter-considerations, and (iii) show how the arguments relate (sub-argument chain, attack, concession, etc.).

3.1 “Health insurance companies should not cover treatment in complementary medicine unless the promised effect and its medical benefit have been concretely proven. Yet this proof is lacking in most cases. Patients do often report relief of their complaints after such treatments. But as long as it is unclear as to how this works, the funds should rather be spent on therapies where one knows with certainty.”

3.2 “There are some programmes on public television that are not worth the licensing fee. Nevertheless, everybody should contribute to the funding of the public broadcasters in equal measure, for we need general and independent media. After all we want to get our view of the world neither through the lens of the government nor through that of rich media entrepreneurs.”

Answer (derived; answers are blank in the source): two arguments per passage, plus one counter-consideration each, as mapped below.

3.1 map

Arg 1   P1 Coverage requires concrete proof of effect and benefit.   ┐
        P2 Such proof is lacking in most cases  ("Yet")              ┴──▶ C1 Insurance should not cover
                                                                              complementary medicine

Arg 2   P3 It is unclear how such treatments work                    ──▶ C2 Funds should rather go to
           (= proof is lacking)  ("But as long as")                          therapies known with certainty

Counter "Patients often report relief"  - - ▶ ⊣ C1   (weakens it, outweighed by "But as long as it is unclear...")

3.2 map

Arg 2   R3 We want our view of the world neither through the government's lens
           nor through rich media owners' lens                ("After all")
                   │
                   ▼
        C2 = R1  We need general and independent media        ("for")
                   │
                   ▼
Arg 1   C1  Everybody should contribute to funding public broadcasters in equal measure

Counter "Some programmes are not worth the fee"  - - ▶ ⊣ C1   (conceded, dismissed by "Nevertheless")

Standard forms

3.1  Arg 1: (1) Coverage requires concrete proof of effect/benefit.
            (2) Such proof is lacking for complementary medicine in most cases.
            ───────────────────────────────────────────────────────────────
            ∴ Health insurance should not cover complementary medicine.

     Arg 2: (1) It is unclear how such treatments work (= proof is lacking).
            ───────────────────────────────────────────────────────────────
            ∴ Funds should rather be spent on therapies where one knows with certainty.

3.2  Arg 1: (1) We need general and independent media.
            ───────────────────────────────────────────────────────────────
            ∴ Everybody should contribute to funding public broadcasters in equal measure.

     Arg 2: (1) We want our view of the world neither through the government's lens
                nor rich media entrepreneurs' lens.
            ───────────────────────────────────────────────────────────────
            ∴ We need general and independent media.   [sub-argument: this conclusion = premise (1) of Arg 1]

Worked solution

Scan for indicators
- Premise indicators: “for”, “after all”, “since”.
- Concession indicators: “nevertheless”, “but”, “yet”: an objection the author acknowledges but does not accept.
Find the main conclusion first
- 3.1: the opening sentence states it (“should not cover … unless proven”).
- 3.2: “everybody should contribute”, flagged by “nevertheless” (despite the concession, here is my claim).
Find the sub-arguments
- 3.2: “for we need general and independent media” gives a premise.
- That premise is itself supported by the “after all” reason: Arg 2 feeds Arg 1 as a chain.
Label the counter-considerations
- 3.1: “patients report relief” argues against the conclusion, outweighed by “but as long as it is unclear…“.
- 3.2: “some programmes are not worth the fee” is conceded and dismissed by “nevertheless”.
Write each argument in standard form
- Premises first, then conclusion; mark which premises come from sub-arguments.

Q4: Syllogism validity

“No P is M. Some S is M. ∴ Some S is not P.”

Answer: Deductive (valid). This is the same argument as mt2 Q6 above and mt3 Q7 below; all three copies share one derivation.

Terms: M = middle term, S = subject of conclusion, P = predicate of conclusion (Example 2.1.1 in Section 2 of the theory file).

0. blank                    1. P1: No P is M            2. P2: Some S is M
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
| S      |     |      P |   | S      |     |      P |   | S      |     |      P |
|   1    |  2  |   3    |   |   1    |  2  |   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+-----+---+    |
|   | 4  |  6  | 5 |    |   |   | 4  |##6##|#5#|    |   |   | 4  |##6##|#5#|    |
|   |    |     |   |    |   |   |    |#####|###|    |   |   | *  |#####|###|    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |      7       |            |      7       |
    | M            |            | M            |            | M            |
    +--------------+            +--------------+            +--------------+

 P1 No P is M    : P and M overlap is empty     -> 5, 6 shaded
 P2 Some S is M  : star in 4 or 6; 6 is empty   -> star in 4
 Conclusion: Some S is not P needs a star in region 1 or 4.   Found: star in 4.

Verdict: valid

Worked solution

Assign terms
- M middle, S subject, P predicate.
Diagram premise 1 (universal, first)
- “No P is M” empties regions 5 and 6.
Diagram premise 2 (particular)
- “Some S is M” needs a star in region 4 or 6.
- Region 6 is already empty, so the star is forced into region 4.
Read off the conclusion
- “Some S is not P” holds when there is a star in region 1 or 4.
- The star landed in 4: the syllogism is valid.

Q5: Syllogism validity

“All M is P. No S is M. ∴ No S is P.”

Answer: Not deductive (invalid). This is Example 2.1.2 in Section 2 of the theory file.

Terms: M = middle term, S = subject of conclusion, P = predicate of conclusion.

0. blank                    1. P1: All M is P           2. P2: No S is M
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
| S      |     |      P |   | S      |     |      P |   | S      |     |      P |
|   1    |  2  |   3    |   |   1    |  2  |   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+-----+---+    |
|   | 4  |  6  | 5 |    |   |   |#4##|  6  | 5 |    |   |   |#4##|##6##| 5 |    |
|   |    |     |   |    |   |   |####|     |   |    |   |   |####|#####|   |    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |######7#######|            |######7#######|
    | M            |            |#M############|            |#M############|
    +--------------+            +--------------+            +--------------+

 P1 All M is P : M outside P is empty      -> 4, 7 shaded
 P2 No S is M  : S and M overlap is empty  -> 4, 6 shaded (4 already)
 Conclusion: No S is P needs the S and P overlap empty = regions 2 and 6.
 Region 6 is shaded, region 2 is still open.

Verdict: invalid

Worked solution

Assign terms
- M middle, S subject, P predicate.
Diagram premise 1
- “All M is P” empties M outside P: regions 4 and 7.
Diagram premise 2
- “No S is M” empties the S and M overlap: regions 4 and 6 (4 already empty).
Check the conclusion
- “No S is P” needs regions 2 and 6 both empty.
- Region 6 is shaded, but region 2 is untouched, so the conclusion is not forced.
- Intuition: an S that is P and outside M is still allowed.

Q6: Syllogism validity

“Some P is M. Some S is not M. ∴ Some S is P.”

Answer: Not deductive (invalid).

Terms: M = middle term, S = subject of conclusion, P = predicate of conclusion.

0. blank                    1. P1: Some P is M          2. P2: Some S is not M
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
| S      |     |      P |   | S      |     |      P |   | S      *     |      P |
|   1    |  2  |   3    |   |   1    |  2  |   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+-----+---+    |
|   | 4  |  6  | 5 |    |   |   | 4  |  6  | 5 |    |   |   | 4  |  6  | 5 |    |
|   |    |     |   |    |   |   |    |     *   |    |   |   |    |     *   |    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |      7       |            |      7       |
    | M            |            | M            |            | M            |
    +--------------+            +--------------+            +--------------+

 P1 Some P is M      : star in 5 or 6, nothing rules either out -> star on the 5/6 line
 P2 Some S is not M  : star in 1 or 2, nothing rules either out -> star on the 1/2 line
 Conclusion: Some S is P needs a star surely in region 2 or 6.
 Both stars sit on lines, so either could be in 5 or 1 instead.  Not forced.

Verdict: invalid

Worked solution

Assign terms
- M middle, S subject, P predicate.
Note that both premises are particular
- There is no universal premise to shade first.
- Border-line rule (theory file): when a star could fall in either of two regions and no universal premise has closed one off, draw it on the line between them (its region is left undetermined).
Diagram premise 1
- “Some P is M” needs a star in region 5 or 6.
- Nothing rules either out, so the star sits on the 5/6 line.
Diagram premise 2
- “Some S is not M” needs a star in region 1 or 2.
- Nothing rules either out, so the star sits on the 1/2 line.
Check the conclusion
- “Some S is P” needs a certain star in region 2 or 6.
- Both stars are on lines and could be in 5 or 1, so nothing is forced: invalid.
- This matches the classical rule that two particular premises never give a valid syllogism.

Q7: Syllogism validity

“No P is M. Some S is M. ∴ Some S is not P.”

This is the same argument as mt2 Q6 and mt3 Q4 above (the source lists it three times across two practice files).

Answer: Deductive (valid). See mt3 Q4’s derivation directly above; it applies unchanged.

0. blank                    1. P1: No P is M            2. P2: Some S is M
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
| S      |     |      P |   | S      |     |      P |   | S      |     |      P |
|   1    |  2  |   3    |   |   1    |  2  |   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+-----+---+    |
|   | 4  |  6  | 5 |    |   |   | 4  |##6##|#5#|    |   |   | 4  |##6##|#5#|    |
|   |    |     |   |    |   |   |    |#####|###|    |   |   | *  |#####|###|    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |      7       |            |      7       |
    | M            |            | M            |            | M            |
    +--------------+            +--------------+            +--------------+

 Conclusion: Some S is not P needs a star in region 1 or 4.   Found: star in 4.

Verdict: valid

mt5 (mid term practice 5.pdf)

Q1 (3 marks): Which sentence is an argument?

(a) Charles went bald, and most men go bald. (b) Charles went bald, like most men. (c) I have been busy since Tuesday. (d) I am busy, since my teacher assigned lots of homework. (e) My roommate likes to ski, and so do I.

Answer: (d).

Verdict table

Item	Argument?	Deciding feature
(a)	no	two facts joined by “and”
(b)	no	”like” introduces a comparison, not a reason
(c)	no	”since Tuesday” marks time, not a reason
(d)	yes	”since” = because: reason for “I am busy”
(e)	no	”and so do I” is additive

(d)   R "my teacher assigned lots of homework"  ──▶  C "I am busy"      ("since")

Worked solution

Check for a premise-conclusion structure
- An argument needs at least one statement offered as a reason for another.
Test (a), (b), (e)
- (a) and (e) only add facts; (b) only compares. None gives a reason.
Test (c)
- “Since Tuesday” means “from Tuesday onward”: a time word.
Test (d)
- “Since” is here a causal/inferential reason indicator.
- Premise: the homework was assigned. Conclusion: I am busy.

Q2 (3 marks): Which sentence is NOT an argument?

(a) He apologized, so you should forgive him. (b) He apologized. Accordingly, you should forgive him. (c) Since he apologized, you should forgive him. (d) If he apologized, you should forgive him. (e) He apologized. As a result, you should forgive him.

Answer: (d).

Verdict table

Item	Argument?	Deciding feature
(a)	yes	”so”
(b)	yes	”Accordingly”
(c)	yes	”Since”
(d)	no	”If” does not assert that he apologized
(e)	yes	”As a result”

Worked solution

Spot what separates (d)
- (a), (b), (c), (e) all assert “he apologized” as a fact and draw a conclusion.
- (d) only says “if” he did.
Why a conditional is not an argument
- “If P then Q” makes no categorical assertion of P.
- With P never claimed true, there is no premise to infer from.
Confirm the others
- Each has an explicit inference indicator and asserts the premise.
- Each then concludes “you should forgive him”.
Conclude
- Only (d) lacks an asserted premise.

Q3 (3 marks): Odd one out

(a) I am tall so I am good at sport. (b) I am tall, as a result I am good at sport. (c) Because I am tall, I am good at sport. (d) I am tall since I am good at sport. (e) I am good at sport since I am tall.

Answer: (e), per verified key.

Note (check): by premise/conclusion roles, (d) is the outlier (there “good at sport” is the premise); (a), (b), (c) and (e) all have premise = “tall”. The key’s (e) rests on surface order only. The key answer is kept.

Roles and order

Item	Premise	Conclusion	Surface order
(a)	I am tall	good at sport	premise first (“so”)
(b)	I am tall	good at sport	premise first (“as a result”)
(c)	I am tall	good at sport	premise first (“Because”)
(d)	good at sport	I am tall	conclusion first (“since”): roles reversed
(e)	I am tall	good at sport	conclusion first (“since”)

Worked solution

Find premise and conclusion in each sentence
- In (a), (b), (c): “I am tall” is the premise, stated before the conclusion.
Check (d)
- “Since” introduces “I am good at sport” as the reason for “I am tall”.
- Roles are reversed compared with (a) to (c).
Check (e)
- “Since” introduces “I am tall” as the reason for “I am good at sport”.
- Roles are the same as (a) to (c), but the conclusion comes first.
Name the pattern difference
- In (a), (b), (c): premise first, conclusion second.
- In (e): conclusion first, premise second.
- The key treats this reversed surface order as the structural odd one out.

Q4 (3 marks): Which argument is deductive?

(a) It flies since it is a bird and birds usually fly. (b) …birds can fly. (c) …I never see a bird that cannot fly. (d) …all birds fly. (e) It flies since it is a bird.

Answer: (d).

Verdict table

Item	Verdict	Deciding feature
(a)	inductive	”usually” admits exceptions
(b)	inductive	”can fly” is a capacity, not “every bird does”
(c)	inductive	generalisation from personal observation
(d)	deductive	”all birds fly” is an unhedged universal
(e)	not valid	no general rule about birds at all

Worked solution

Recall the criterion
- Deductive: if the premises are true, the conclusion cannot be false.
- In natural language the signal is a universal rule with no hedging.
Test (a), (b), (c)
- (a) “usually” leaves room for exceptions.
- (b) “can fly” is a capacity, not a fact about every bird.
- (c) “never see” covers only what the speaker has observed.
Test (d)
- “All birds fly” plus “it is a bird” gives “it flies” by universal instantiation: deductive.
Test (e)
- “It is a bird” alone, with no rule about birds, cannot support “it flies”.
- The linking premise is absent: not a valid argument in any form.

Q5 (3 marks): Which sentence cannot be a conjunction of two simpler sentences?

(a) Miranda and Nick ate at home. (b) Miranda and Nick played tennis. (c) A Catholic priest married John and Mary. (d) Fred had pie and ice cream for dessert. (e) Jane speaks French and English.

Answer: (c).

Verdict table

Item	Splits into two sentences?	Deciding feature
(a)	yes	Miranda ate at home; Nick ate at home
(b)	yes	Miranda played tennis; Nick played tennis (distributive reading)
(c)	no	”married John and Mary” is one joint act on the pair
(d)	yes	Fred had pie; Fred had ice cream
(e)	yes	Jane speaks French; Jane speaks English

Worked solution

Recall what a conjunction of two simpler sentences means
- S splits into “S1 and S2” if S1 and S2 are each meaningful and together say the same as S.
- Formally S is P ∧ Q.
Test (a), (d), (e)
- The two halves are independent claims that together equal the original.
Test (b)
- They may have played together (doubles) or separately.
- The distributive reading (“each played tennis”) works as a conjunction even if the together reading is also possible.
Test (c)
- Split: “a priest married John” and “a priest married Mary” would mean two separate ceremonies.
- The original says the priest married John and Mary to each other, in one ceremony.
- “Married John and Mary” is one relational predicate applied to the pair: it cannot be decomposed.

Q6, Q7, Q8: Which Venn diagram correctly represents the given categorical proposition(s)?

These three questions each present a pre-drawn 3-circle Venn diagram as an image (with shaded regions and/or asterisks already marked) and ask the student to pick the matching answer choice: (a)-(f) for Q6, (a)-(e) for Q7, (a)-(f) for Q8. The diagrams themselves are figures in the source PDF and were not captured by text extraction; only the multiple-choice option text survived.

Answer: cannot be filled in without the source figure (same limitation as mt5 Q13/14 above).

Note (check): “mt5 Q13/14 above” does not resolve inside Part 1 (those questions are not listed here); the wording is kept verbatim. The propositions themselves are not in the file, so no per-question diagram can be redrawn. A generic gallery of the four forms is given below instead.

Reference: how each form looks on the diagram (S and P shown; M is the third circle)

A: All S is P               E: No S is P                I: Some S is P
+--------+-----+--------+   +--------+-----+--------+   +--------+-----+--------+
|#S######|     |      P |   | S      |#####|      P |   | S      |     |      P |
|###1####|  2  |   3    |   |   1    |##2##|   3    |   |   1    |  2  |   3    |
|   +----+-----+---+    |   |   +----+-----+---+    |   |   +----+--*--+---+    |
|###|#4##|  6  | 5 |    |   |   | 4  |##6##| 5 |    |   |   | 4  |  6  | 5 |    |
|###|####|     |   |    |   |   |    |#####|   |    |   |   |    |     |   |    |
+---+----+-----+---+----+   +---+----+-----+---+----+   +---+----+-----+---+----+
    |      7       |            |      7       |            |      7       |
    | M            |            | M            |            | M            |
    +--------------+            +--------------+            +--------------+

O: Some S is not P
+--------+-----+--------+
| S      |     |      P |
|   1    |  2  |   3    |
|   +-*--+-----+---+    |
|   | 4  |  6  | 5 |    |
|   |    |     |   |    |
+---+----+-----+---+----+
    |      7       |
    | M            |
    +--------------+

 A  shades S outside P     = regions 1, 4
 E  shades S and P overlap = regions 2, 6
 I  star in S and P        = region 2 or 6 (on the 2/6 line if neither is closed off)
 O  star in S outside P    = region 1 or 4 (on the 1/4 line if neither is closed off)

Worked solution (general method, so the diagrams can be solved once seen)

Identify the proposition(s) behind each answer choice
- Translate every option into standard categorical form (A/E/I/O), using the method from mt1 Q11 and Section 2 of the theory file.
Draw the 7-region diagram for each candidate
- Universal propositions (“All S are P”, “No S are P”) are shading.
- Particular propositions (“Some S are P”, “Some S are not P”) are a star, drawn after any universal premise has been shaded.
Compare with the given figure region by region
- The right choice matches shading and star placement exactly.
- Check all 7 regions: an unshaded region carries information too (it means “not asserted to be empty”).
Watch for the two classic traps
- (i) A universal drawn as a star (or the reverse) is always wrong: shading means empty, a star means at least one exists.
- (ii) A star drawn squarely inside one region when the premise is ambiguous between two is wrong, unless a universal premise has already closed the other region (see the border-line rule in mt3 Q6).

mt6 (mid term practice 6.pdf)

Q1 (10 marks): Does the proposition have the form ¬S?

For each sentence, state whether it can be read as the propositional negation ¬S of some simpler proposition S. If yes, identify S.

(a) “You may not go to the meeting.” (b) “I cannot recommend him too highly.” (c) “He never thought he’d go to the Himalayas.” (d) “I have not done all questions.” (e) “No smoking section available.”

Answer:

(a) Yes; S = “you are allowed to go to the meeting.”

(b) Yes (under the low-recommendation reading); S = “I can recommend him highly.”

(c) Yes; S = “He thought he’d go to the Himalayas.”

(d) Yes; S = “I have done all questions.”

(e) Yes (under the “there is no smoking section” reading); S = “there is a smoking section.”

Sentence	Keyword trap	S	Form
(a) You may not go to the meeting	”may not” = not allowed	you are allowed to go to the meeting	¬S
(b) I cannot recommend him too highly	double-negation idiom; only the low reading gives ¬S	I can recommend him highly	¬S (low reading)
(c) He never thought he’d go to the Himalayas	”never” = not ever	he thought he’d go to the Himalayas	¬S
(d) I have not done all questions	”not all” is not “none”	I have done all questions	¬S
(e) No smoking section available	”no” = there is not	there is a smoking section	¬S (this reading)

Note (check): in (b) the idiomatic reading of “cannot … too highly” is an emphatic positive recommendation, which is not a plain ¬S. The key’s answer holds only under the low reading. The key is kept.

Worked solution

Understand the task
- Find an S such that “not S” accurately paraphrases the sentence under at least one natural reading.
Watch where negation scopes
- “Not all” does not mean “none”; “never” means “not ever”.
- “Cannot … too” is a double-negation idiom.
- Goal: isolate a clean S and its plain negation.
Apply to (a) and (c)
- (a) “You may not go” = “it is not the case that you are allowed to go”.
- (c) “He never thought he’d go” = “it is not the case that he thought he’d go”.
Apply to (b)
- Idiomatic reading: “no level of praise is too high”, effectively an emphatic recommendation.
- Low reading: “I am not able to recommend him highly”, so S = “I can recommend him highly” and the sentence is ¬S.
Apply to (d) and (e)
- (d) is straightforward: ¬(I have done all questions).
- (e) reads as “there is no smoking section” = ¬(there is a smoking section).

Q2 (10 marks): Best categorical form for NL propositions

Restate each sentence in the best standard categorical form (A/E/I/O) and identify the subject class S and predicate class P.

(a) “My friends are the only ones who care.” (b) “Everything that is cheap is no good.” (c) “Some things that are cheap are not good.” (d) “Only seniors may take this course.” (e) “Somebody does not love you.”

Answer:

(a) “No S are P” where S = people who are not my friends, P = people who care.

(b) “All S are P” (S = cheap things, P = no-good things); equivalently “No S are P” (S = cheap things, P = good things).

(c) “Some S are not P” where S = cheap things, P = good things (or equivalently “Some S are P” where P = no-good things).

(d) “All S are P” (S = students who may take this course, P = seniors); equivalently “No S are P” (S = non-seniors, P = students who may take this course).

(e) “Some S are not P” where S = people, P = you-lovers.

(The cheat sheet of NL patterns to A/E/I/O is at mt1 Q11.)

Sentence	Keyword trap	S	P	Form
(a) My friends are the only ones who care	”the only ones” swaps the terms	people who are not my friends	people who care	E
(b) Everything that is cheap is no good	”no good” is a negated predicate, not an E	cheap things	no-good things	A (or E with P = good things)
(c) Some things that are cheap are not good	”some … not”	cheap things	good things	O (or I with P = no-good things)
(d) Only seniors may take this course	”only” swaps the terms	students who may take this course	seniors	A (or E: non-seniors / course-takers)
(e) Somebody does not love you	”somebody … not”	people	you-lovers	O

Worked solution

Recall the four forms
- A: All S are P. E: No S are P. I: Some S are P. O: Some S are not P.
Handle “only” and “the only ones”
- These invert the natural reading: “Only S are P” means “All P are S”.
- “Only seniors may take it” gives “all who may take it are seniors”.
- “My friends are the only ones who care” gives “all who care are my friends”, equivalently “no non-friend cares”.
Apply to (a) and (d)
- (a) Pick S = non-friends, P = carers: “No S are P”.
- (d) S = those who may take the course, P = seniors: “All S are P”.
Apply to (b) and (c)
- (b) is a universal: All cheap things are no-good things; using the complement of P, No cheap things are good things.
- (c) is a particular negative: Some S are not P (P = good things), or Some S are P with P = no-good things.
Apply to (e)
- “Somebody does not love you” means there is a person who is not a you-lover.
- S = people, P = you-lovers: “Some S are not P”.

Q3 (10 marks): Suppressed premises for deductive validity

Supply the missing premise(s) needed to make each argument deductively valid. Rewrite in standard form where helpful.

(a) “A fetus is a human being and hence abortion is illegal.” (b) A student writes a proof of the formula 0 + 1 + 2 + … + n = n(n+1)/2 by mathematical induction: they show it holds for n=0 (base case), then show that if it holds for m it holds for m+1 (inductive step), and conclude it holds for all natural numbers n. What is the suppressed premise? (c) “This formal language is not regular, since otherwise {aⁿbⁿ | n∈N} would be regular as well.”

Answer:

(a) two suppressed premises: “Abortion kills a fetus” and “Killing a human being is illegal”.

(b) the Mathematical Induction Principle: “For any property P, if P holds for 0, and if P holds for m then P holds for m+1, then P holds for every natural number.”

(c) two suppressed premises: (1) “If this language is regular, then {aⁿbⁿ | n∈N} is also regular” and (2) “{aⁿbⁿ | n∈N} is not regular.”

(a)   A fetus is a human being                     [stated]
      Abortion kills a fetus                       [suppressed]
      ─────────────────────────────────────
      ∴ Abortion kills a human being
      Killing a human being is illegal             [suppressed]
      ─────────────────────────────────────
      ∴ Abortion is illegal

(b)   P holds for 0                                [base case, stated]
      P(m) implies P(m+1) for all m                [inductive step, stated]
      Mathematical Induction Principle             [suppressed]
      ─────────────────────────────────────
      ∴ P holds for every natural number n

(c)   If this language is regular, then {aⁿbⁿ | n∈N} is regular    [suppressed]
      {aⁿbⁿ | n∈N} is not regular                                  [suppressed]
      ──────────────────────────────────────────────────────────
      ∴ This formal language is not regular

Note (check): in (a) “Killing a human being is illegal” is not true without exceptions (self-defence, war), so the argument is valid but that premise is contestable. Kept as in the key.

Note (check): in (c) the stated clause “otherwise … would be regular” already expresses premise (1), so arguably only premise (2) is truly suppressed. The key lists both. Kept as in the key.

Worked solution

Recall the method for suppressed premises
- Ask what general rule or background fact, combined with the stated premises, makes the inference deductively valid.
Apply to (a)
- The stated premise is only “a fetus is a human being”.
- Two bridges are needed: abortion kills a fetus, and killing a human being is illegal.
- Both are assumed without being stated.
Apply to (b)
- The proof gives a base case and an inductive step.
- Without the Induction Principle itself, going from those to “holds for all n” is not justified.
- The principle is the suppressed premise.
Apply to (c)
- The stated premise is “otherwise {aⁿbⁿ} would be regular”.
- Two facts are needed: if the language is regular then {aⁿbⁿ} is regular (a reduction/closure property), and {aⁿbⁿ} is not regular (the Pumping Lemma result).
- Both are standard results in formal language theory and are left unstated.

Q4 (10 marks): Form arguments and identify attacks

KB: r1: Alice, Bob, or Cr is not innocent (disjunction). r2: presumption of innocence if not proven guilty. r3: Alice not proven guilty. r4: Bob not proven guilty. r5: newborns always innocent. r6: Cr is a newborn.

C  = r5, r6           ∴ Cr is innocent
A  = r3, r2            ∴ Alice is presumably innocent
B  = r4, r2            ∴ Bob is presumably innocent
A' = r1, B, C          ∴ Alice is not innocent
B' = r1, A, C          ∴ Bob is not innocent

Answer: A and A’ rebut each other (contradictory conclusions: “Alice is innocent” vs “Alice is not innocent”); B and B’ rebut each other likewise.

Argument map with attacks

 r5, r6 ──▶ C   Cr is innocent
 r3, r2 ──▶ A   Alice is presumably innocent  ◀───── rebut ─────▶  A'   Alice is not innocent
 r4, r2 ──▶ B   Bob is presumably innocent    ◀───── rebut ─────▶  B'   Bob is not innocent

 A' = r1 + B + C ──▶ Alice is not innocent      (uses B and C as sub-arguments)
 B' = r1 + A + C ──▶ Bob is not innocent        (uses A and C as sub-arguments)

Attacker	Target	Type	Why
A	A’	rebut	”Alice innocent” vs “Alice not innocent”
A’	A	rebut	same pair, other direction
B	B’	rebut	”Bob innocent” vs “Bob not innocent”
B’	B	rebut	same pair, other direction

Note (check): A’ contains B as a sub-argument, and B’ concludes “Bob is not innocent”, so B’ also attacks A’ through its sub-argument B; symmetrically A’ attacks B’ through A. Under a sub-argument-aware reading these extra attacks exist. The key lists only the four above; it is kept as given.

Worked solution

Recall the three types of attack
- Rebut: two arguments have directly contradictory conclusions (P and not-P).
- Undermine: attacks a premise of the target (claims a stated premise is false).
- Undercut: attacks the inferential link itself (the rule does not apply here), without contradicting premise or conclusion.
Identify the conclusions
- A concludes: Alice is (presumably) innocent.
- A’ concludes: Alice is not innocent.
- One is the negation of the other.
Classify A versus A’
- Contradictory conclusions mean each attacks the other directly.
- Rebutting attack in both directions: (A, A’) and (A’, A).
Apply the same analysis to B and B’
- B concludes Bob is innocent; B’ concludes Bob is not innocent.
- So (B, B’) and (B’, B) are rebutting attacks.
Check for other attack types
- No argument here attacks a premise of another without that premise being a conclusion of a sub-argument.
- No argument targets a rule, so there are no undercutters in the given list.

Q5 (10 marks): Barber paradox arguments

“A barber shaves all those, and those only, who do not shave themselves.” Argument A assumes the barber shaves himself and derives that he doesn’t; Argument B assumes he doesn’t and derives that he does.

A = (1) If X shaves himself, the barber does not shave X.
    (2) The barber shaves himself. (assumption)
    ∴ The barber does not shave the barber.
B = (1) If X does not shave himself, the barber shaves X.
    (2) The barber does not shave himself. (assumption)
    ∴ The barber shaves the barber.

Answer: Both A and B are deductive. A and B rebut each other (contradictory conclusions). A also undermines itself, since its conclusion contradicts its own premise (2); likewise B undermines itself. So the framework has four attacks: (A,A), (A,B), (B,B), (B,A).

Argument map with attacks

 A:  assume barber shaves himself      ──▶  barber does not shave the barber
 B:  assume barber does not shave self ──▶  barber shaves the barber

        rebut (contradictory conclusions)
     ┌────────────────────────────────────┐
     ▼                                    │
    A  ◀──────────────────────────────▶  B
   ⟲                                       ⟲
   self-undermine                         self-undermine
   (concl. contradicts own premise 2)     (concl. contradicts own premise 2)

 Att = { (A,A), (A,B), (B,B), (B,A) }

Worked solution

Verify both arguments are deductive
- A uses a universal rule plus the assumption about the barber; instantiating X = barber gives the conclusion.
- B has the same structure with the negated assumption.
- Both are valid modus ponens forms.
Find the rebutting attacks between A and B
- A concludes “the barber does not shave the barber”; B concludes “the barber shaves the barber”.
- Direct contradiction, so both (A, B) and (B, A) hold.
Find the self-undermining in A
- Undermining attacks a premise of an argument.
- A’s conclusion “the barber does not shave the barber” contradicts A’s own premise (2), “the barber shaves himself”.
- So (A, A).
Apply the same reasoning to B
- B’s conclusion “the barber shaves the barber” contradicts its own premise (2), “the barber does not shave himself”.
- So (B, B).
Collect all four attacks
- Att = {(A,A), (A,B), (B,B), (B,A)}.
- This four-attack pattern is the signature of the barber paradox: neither assumption yields a stable, self-consistent argument.

Part 2: Computational Argumentation

How to read the diagrams

A ──▶ B        A attacks B
A ◀──▶ B       A and B attack each other
A ↺            A attacks itself
[A]            in the extension
(A)            out: attacked by something in the extension
?A?            undecided: neither in nor out
f^i            i-th stage of the grounded iteration, f^0 = the empty set

Every finite AF below was checked by a script (conflict-free, admissible, preferred, stable, grounded); all claims in this part matched.

mt5 (mid term practice 5.pdf)

Q9 (5 marks): full mutual 3-cycle

F=(AR,Att), AR={A1,A2,A3}, Att={(A1,A2),(A1,A3),(A2,A1),(A2,A3),(A3,A1),(A3,A2)} (every argument attacks the other two).

Which statement is not correct? (a) No arguments are groundedly acceptable. (b) A1,A2,A3 are all admissibly acceptable. (c) There are three admissible extensions. (d) ∅ is an admissible extension. (e) ∅ is the grounded extension.

Answer: (c)

Graph (every side is a two-way attack)

            A1
          ◀╱  ╲▶
          ╱    ╲
        A2 ◀────▶ A3

Final labellings

   preferred {A1}      preferred {A2}      preferred {A3}      grounded (GE = {})
       [A1]                (A1)                (A1)                ?A1?
      ◀╱  ╲▶              ◀╱  ╲▶              ◀╱  ╲▶              ◀╱  ╲▶
   (A2) ◀▶ (A3)        [A2] ◀▶ (A3)        (A2) ◀▶ [A3]        ?A2? ◀▶ ?A3?

Worked solution

Find all conflict-free sets
- A set is conflict-free if no member attacks another member.
- Every pair attacks each other, so no 2-element set and not the full set is conflict-free.
- No argument attacks itself, so ∅ and the three singletons are conflict-free.

Test each set for admissibility

set                conflict-free?  admissible?  reason
{}                 yes             yes          nothing to defend
{A1}               yes             yes          attackers A2, A3 are both attacked by A1
{A2}               yes             yes          attackers A1, A3 are both attacked by A2
{A3}               yes             yes          attackers A1, A2 are both attacked by A3
any pair, triple   no              no           contain an attacking pair

Count the admissible extensions
- ∅ plus three singletons gives four: {∅, {A1}, {A2}, {A3}}.
Identify the false statement
- (c) says three, but there are four. So (c) is false.
Verify the other options with the grounded iteration
```
stage  set   newly defended, and why
f^0    {}    start
f^1    {}    every argument has attackers and {} attacks nothing, so nothing is defended
```
- $GE = \emptyset$ , confirming (a) and (e).
- (d) holds from the table, and (b) holds because each Ai lies in the admissible set {Ai}.

Q10 (5 marks): directed 3-cycle

AR={A1,A2,A3}, Att={(A1,A2),(A2,A3),(A3,A1)}. Which statement is not correct?

Answer: (b)

Graph

      A1 ─────▶ A2
       ▲         │
       │         ▼
       └──────── A3

Final labelling: all undecided, $GE = \emptyset$ .

     ?A1? ─────▶ ?A2?
       ▲            │
       │            ▼
       └───────── ?A3?

Worked solution

Check for self-attacks
- None of A1, A2, A3 attacks itself, so every singleton is conflict-free.

Test each set for admissibility

set        conflict-free?  admissible?  reason
{}         yes             yes          nothing to defend
{A1}       yes             no           attacker A3 is not attacked by A1 (A1 attacks only A2)
{A2}       yes             no           attacker A1 is not attacked by A2 (A2 attacks only A3)
{A3}       yes             no           attacker A2 is not attacked by A3 (A3 attacks only A1)

Each argument’s only attacker is the “previous” one in the cycle, and nobody attacks two steps ahead.

Conclude
- The only admissible extension is ∅, so no argument is admissibly acceptable.
- So option (b) (“A1, A2, A3 are all admissibly acceptable”) is the false statement.
Grounded extension
```
stage  set   newly defended, and why
f^0    {}    start
f^1    {}    every Ai has exactly one attacker and {} attacks nothing
```
- $GE = \emptyset$ , confirming (a), (d), (e): no groundedly acceptable arguments, ∅ is admissible, ∅ is the grounded extension.

Q11

AR={A1,A2,A3,A4}, Att={(A2,A1),(A1,A3),(A2,A2)}. How many arguments are admissibly acceptable? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4

Answer: (b) 1

Graph

    A2 ↺ ───▶ A1 ───▶ A3          A4

Final labelling (extension {A4})

    ?A2? ↺ ───▶ ?A1? ───▶ ?A3?          [A4]

Nothing in the extension attacks A1, A2 or A3, so they stay undecided.

Worked solution

Identify self-attacking arguments
- (A2,A2) is in Att, so A2 attacks itself and no conflict-free set contains A2.
Assess A1
- A1’s only attacker is A2, so a defending set must attack A2.
- Only A2 attacks A2, and A2 is in no conflict-free set. A1 can never be defended.
Assess A3
- A3’s only attacker is A1, so something must attack A1.
- Only A2 attacks A1, and it is excluded. A3 can never be defended.
Assess A4
- A4 has no attackers, so {A4} is conflict-free and vacuously defends A4.

Table of candidate sets

set                conflict-free?  admissible?  reason
{}                 yes             yes          nothing to defend
{A1}               yes             no           A2 attacks A1, nothing in the set attacks A2
{A3}               yes             no           A1 attacks A3, nothing in the set attacks A1
{A4}               yes             yes          unattacked
{A1,A4}, {A3,A4}   yes             no           A1 (resp. A3) is still undefended
{A1,A3}, ...       no              no           A1 attacks A3 (and any set with A2 self-attacks)

Count
- Exactly one argument (A4) is admissibly acceptable: (b) 1.

Q12: infinite backward chain

AR={A0,A1,A2,…}, Att={(Ai+1,Ai) | i=0,1,2,…} (each Ai is attacked by Ai+1). How many admissible extensions does F have? (a) 1 (b) 2 (c) 3 (d) 4 (e) none of the above

Answer: (e)

Graph (first few nodes, then ”…“)

    A0 ◀─── A1 ◀─── A2 ◀─── A3 ◀─── A4 ◀─── ...

Two of the infinitely many admissible sets

   S_0 = {A0, A2, A4, ...}    [A0] ◀─── (A1) ◀─── [A2] ◀─── (A3) ◀─── [A4] ...
   S_1 = {A1, A3, A5, ...}    (A0) ◀─── [A1] ◀─── (A2) ◀─── [A3] ◀─── (A4) ...

Worked solution (derived)

No finite non-empty set is admissible
- Let Ak be the member of S with the highest index k. It is attacked by Ak+1, which is not in S.
- Only Ak+2 attacks Ak+1, and it is not in S either. So Ak is left undefended.
∅ is admissible
- It is conflict-free and defends everything in it, vacuously.
Construct infinitely many infinite admissible sets
- For each n ≥ 0 let Sₙ = {An, An+2, An+4, …} (every other argument from An).
- Members of Sₙ are at least two steps apart, and only consecutive arguments attack each other, so Sₙ is conflict-free.
- Each Ak in Sₙ is attacked only by Ak+1 (not in Sₙ), and Ak+2 ∈ Sₙ attacks Ak+1, so Sₙ defends Ak.
```
set                       admissible?  reason
{}                        yes          nothing to defend
finite non-empty S        no           the top member cannot be defended (step 1)
S_n = {An, An+2, ...}     yes          conflict-free, and Ak+2 answers every attack from Ak+1
```
Count
- One extension ∅, plus one Sₙ for each n = 0, 1, 2, …: countably infinitely many.
- The answer is not 1, 2, 3 or 4, so it is (e).

Grounded extension (for reference):

stage  set   newly defended, and why
f^0    {}    start
f^1    {}    every Ai is attacked by Ai+1, so no argument is unattacked

$GE = \emptyset$ , although A0 is admissibly acceptable (it lies in S_0).

Q13 / Q14: ASPIC+ quarantine(bob)

“Explain in natural language what knowledge is represented by the following ASPIC+ argumentation theory. Explain whether one should accept the proposition quarantine(bob) by constructing possible arguments and evaluate their acceptabilities.”

Answer: (the underlying strict/defeasible rule set was presented as a diagram/image in the PDF and was not captured by text extraction: it cannot be filled in without the source figure)

The general method

 knowledge base (axioms, strict rules, defeasible rules)
        │  build arguments by chaining rules (sub-arguments count too)
        ▼
 arguments AR
        │  find rebut / undermine / undercut
        ▼
 attacks Att
        │  form the framework (AR, Att)
        ▼
 compute an extension (grounded / preferred / admissible)
        │
        ▼
 is there an accepted argument whose conclusion is quarantine(bob)?

The three attack types

 rebut       X: ... ⇒ p              Y: ... ⇒ not p           X ◀──rebut──▶ Y
             (conclusions contradict; the attacked argument's top rule is defeasible)

 undermine   X: ... ⇒ not q          Y: uses defeasible premise q
                                                              X ──undermine──▶ Y

 undercut    X: ... ⇒ "rule d does not apply"     Y: uses defeasible rule d
                                                              X ──undercut──▶ Y

Worked solution (general method)

Read the knowledge base
- Strict rules cannot be attacked, defeasible rules can be defeated, axioms are facts.
- Read the diagram to identify each rule’s type, premises and conclusion.
Build all arguments
- Forward-chain from the axioms through the rules. Partial chains (sub-arguments) are arguments too.
Identify attacks
- Rebut: contradictory conclusions, with a defeasible last step in the attacked argument.
- Undermine: the conclusion contradicts a defeasible premise used by another argument.
- Undercut: the conclusion says “rule d does not apply”, targeting a defeasible rule.
Form the framework (AR, Att)
- Collect all arguments as AR and all attacks as Att.
Apply semantics
- Compute the grounded (or preferred/admissible) extension.
- Check whether an argument with conclusion quarantine(bob) is in it.
Report
- In the grounded extension: groundedly acceptable.
- Only in some preferred extension: admissibly but not groundedly acceptable.
- In none: not acceptable.

Q15: GE is admissible (proof)

“Let F=(AR,Att) be an argumentation framework; f:2^AR→2^AR where f(S)={X∈AR | S defends X}; and GE = f(∅) ∪ f²(∅) ∪ f³(∅) ∪ … . Prove that GE is an admissible extension of F.”

Answer: GE is conflict-free and self-defending, hence admissible.

Proof idea: the stages grow, so any two members of GE sit in one stage. A clash inside stage $i + 1$ would force a clash inside stage $i$ , so every stage is conflict-free. Each member was defended by an earlier stage, which lies inside GE.

The stage chain

   ∅ = f^0(∅)  ⊆  f^1(∅)  ⊆  f^2(∅)  ⊆  ...        GE = union of all stages

Picture of the clash argument (X, Y both in stage i+1, X attacks Y)

   X ──▶ Y          both in f^(i+1)
   ▲
   Z                Z in f^i attacks X, because f^i defends Y
   ▲
   W                W in f^i attacks Z, because f^i defends X
                    W, Z in f^i with W ──▶ Z: f^i is not conflict-free

Proof

Part 1: the chain
 1. f is monotone: S ⊆ T implies f(S) ⊆ f(T)   because a bigger set attacks at least as much
 2. f^0(∅) ⊆ f^1(∅) ⊆ f^2(∅) ⊆ ...           because ∅ ⊆ f(∅) and monotonicity, by induction
 3. GE is well defined, and any two members lie in one common stage   because the stages are nested

Part 2: GE is conflict-free (induction: Q(i) says f^i(∅) is conflict-free)
 4. Q(0) holds                                   because f^0(∅) = ∅ has no elements
 5. Assume Q(i). Suppose X, Y ∈ f^(i+1)(∅) and X attacks Y   (for contradiction)
 6. Some Z ∈ f^i(∅) attacks X                    because f^i(∅) defends Y and X attacks Y
 7. Some W ∈ f^i(∅) attacks Z                    because f^i(∅) defends X and Z attacks X
 8. W, Z ∈ f^i(∅) with W attacking Z             contradicts Q(i)
 9. Q(i+1) holds
10. GE is conflict-free                          because any two members lie in one conflict-free stage

Part 3: GE defends all its members
11. Take X ∈ GE. Then X ∈ f^i(∅) for some i ≥ 1
12. f^(i-1)(∅) defends X                         because X ∈ f(f^(i-1)(∅))
13. So every attacker Y of X is attacked by some Z ∈ f^(i-1)(∅)
14. Z ∈ GE                                       because f^(i-1)(∅) ⊆ GE
15. GE defends X

Conclusion
16. GE is conflict-free (10) and defends every member (15), so GE is admissible.        ∎

mt4 (mid term practice 4.pdf, answer-key excerpt)

Q17: two arguments A, B (statement truncated in source)

Answer:

conflict-free = {∅,{A},{B}}

admissible = {∅,{A},{B}}

GE = ∅

admissibly accepted: A, B; groundedly accepted: none.

Graph (reconstructed: Att = {(A,B),(B,A)})

    A ◀──────▶ B

Final labellings

   extension {A}       extension {B}       grounded (GE = {})
   [A] ◀──▶ (B)        (A) ◀──▶ [B]        ?A? ◀──▶ ?B?

Worked solution

Reconstruct the framework
- The answer fits a mutual-attack framework: Att = {(A,B),(B,A)}, no other attacks.

Test each set

set     conflict-free?  admissible?  reason
{}      yes             yes          nothing to defend
{A}     yes             yes          B attacks A, and A attacks B back
{B}     yes             yes          A attacks B, and B attacks A back
{A,B}   no              no           A attacks B

Grounded extension

stage  set   newly defended, and why
f^0    {}    start
f^1    {}    A and B both have attackers, and {} attacks nothing

State acceptability
- Admissible semantics: A ∈ {A} and B ∈ {B}, so both are admissibly acceptable.
- Grounded semantics: GE = ∅, so neither is groundedly acceptable.

Q18: AR={A0,A1,A2}, Att={(A1,A0),(A2,A1),(A0,A2),(A2,A2)}

Answer:

conflict-free = {∅,{A0},{A1}}

admissible = {∅}

GE = ∅

nobody is admissibly or groundedly accepted.

Graph

      A0 ────────▶ A2 ↺
       ▲            │
       │            ▼
       └─────────── A1

Final labelling: all undecided, $GE = \emptyset$ .

     ?A0? ───────▶ ?A2? ↺
       ▲              │
       │              ▼
       └───────────?A1?

Worked solution

Identify self-attacks
- (A2,A2) ∈ Att, so any set containing A2 is not conflict-free.

Find conflict-free sets and test admissibility

set        conflict-free?  admissible?  reason
{}         yes             yes          nothing to defend
{A0}       yes             no           attacker A1; only A2 attacks A1, and A2 is in no conflict-free set
{A1}       yes             no           attacker A2; only A0 attacks A2, and A0 is not in the set
{A0,A1}    no              no           A1 attacks A0
with A2    no              no           self-attack

Note on ${A_{1}}$
- A2 self-attacks but is still an attacker of A1, so A1 still needs a defender.

Grounded extension

stage  set   newly defended, and why
f^0    {}    start
f^1    {}    A0 is attacked by A1, A1 by A2, A2 by A0 and itself: nobody is unattacked

Conclude
- Only ∅ is admissible and $GE = \emptyset$ : no argument is admissibly or groundedly acceptable.

Q19: AR={A0,A1,A2,A3}, Att={(A1,A0),(A2,A1),(A0,A2),(A3,A2)}

Answer:

conflict-free = {∅,{A0},{A1},{A2},{A3},{A0,A3},{A1,A3}}

admissible = {∅,{A3},{A1,A3}}

GE={A1,A3}

admissibly and groundedly accepted: A1, A3.

Graph

      A0 ───────▶ A2 ◀─────── A3
       ▲           │
       │           ▼
       └────────── A1

Final labelling (extension {A1,A3})

     (A0) ───────▶ (A2) ◀─────── [A3]
       ▲             │
       │             ▼
       └──────────  [A1]

A2 is out (A3 attacks it), so its attack on A1 no longer counts. A0 is out (A1 attacks it).

Worked solution

Map the attacks
- A1 attacks A0. A2 attacks A1. A0 attacks A2. A3 attacks A2. No self-attacks.
- Attackers: A0 by A1; A1 by A2; A2 by A0 and A3; A3 by nobody.
Find conflict-free sets
- Four pairs conflict: {A0,A1}, {A1,A2}, {A0,A2}, {A2,A3}.
- {A0,A3} and {A1,A3} survive. Any triple or the full set contains one of the four conflicting pairs.

Test each set for admissibility

set        conflict-free?  admissible?  reason
{}         yes             yes          nothing to defend
{A0}       yes             no           A1 attacks A0; A0 attacks only A2
{A1}       yes             no           A2 attacks A1; nothing in {A1} attacks A2
{A2}       yes             no           A0 and A3 attack A2; A2 attacks neither
{A3}       yes             yes          unattacked
{A0,A3}    yes             no           A0 still undefended (A3 does not attack A1)
{A1,A3}    yes             yes          A3 attacks A2, so A1 is defended; A3 is unattacked

Grounded extension by iteration

stage  set      newly defended, and why
f^0    {}       start
f^1    {A3}     A3 has no attacker
f^2    {A1,A3}  A1's attacker A2 is attacked by A3, so A1 is defended.
                A0 stays out (A1 is not attacked by {A3}).
                A2 stays out: A3 attacks A2, which is the opposite of defending it.
f^3    {A1,A3}  A0 needs an attack on A1, A2 needs attacks on A0 and A3: none. Fixed point.

$GE = {A_{1}, A_{3}}$ .

Q20: AR={A0,A1,A2}, Att={(A1,A0),(A2,A0),(A1,A2),(A2,A1)} (“double cycle”)

Answer:

conflict-free = {∅,{A0},{A1},{A2}}

admissible = {∅,{A1},{A2}}

GE = ∅

admissibly accepted: A1, A2; groundedly accepted: none.

Graph

      A1 ◀───────▶ A2
        ╲          ╱
         ╲        ╱
          ▼      ▼
            A0

Final labellings

   extension {A1}         extension {A2}         grounded (GE = {})
   [A1] ◀──▶ (A2)         (A1) ◀──▶ [A2]         ?A1? ◀──▶ ?A2?
     ╲       ╱              ╲       ╱              ╲        ╱
      ▼     ▼                ▼     ▼                ▼      ▼
        (A0)                   (A0)                   ?A0?

Worked solution

Map the attacks
- A1 attacks A0 and A2. A2 attacks A0 and A1. A0 attacks nobody.

Find conflict-free sets and test admissibility

set        conflict-free?  admissible?  reason
{}         yes             yes          nothing to defend
{A0}       yes             no           attacked by A1 and A2; A0 attacks nobody
{A1}       yes             yes          A2 attacks A1, and A1 attacks A2 back
{A2}       yes             yes          A1 attacks A2, and A2 attacks A1 back
{A1,A2}    no              no           mutual attack
{A0,A1}    no              no           A1 attacks A0
{A0,A2}    no              no           A2 attacks A0

Grounded extension

stage  set   newly defended, and why
f^0    {}    start
f^1    {}    A0 is attacked by A1 and A2, A1 by A2, A2 by A1: nobody is unattacked

$GE = \emptyset$ : no argument is groundedly acceptable.

mt7 (mid term practice 7.pdf)

Q5: knowledge base to arguments

KB:

d1 “Students are normally not married”
d2 “Students are normally young adults”
s1 “Young adults are adults”
d3 “Adults are normally married”
e1 “Alex is a student”

(a) Form all arguments A1={e1,d1 ∴ Alex is not married}, A2={e1,d2 ∴ Alex is a young adult}, A3={A2,s1 ∴ Alex is an adult}, A4={A3,d3 ∴ Alex is married}. (b) Identify attack relationships (rebutting/undermining/undercutting).

Answer: A1 and A4 rebut each other (contradictory conclusions “not married” vs “married”).

The arguments (axiom at the leaf, conclusion at the root)

 A1                        A4
 Alex is not married       Alex is married          (top rule d3, defeasible)
        ▲                        ▲
        │ d1 (defeasible)        │ d3 (defeasible)
 e1: Alex is a student     A3: Alex is an adult     (top rule s1, strict)
     (axiom)                     ▲
                                 │ s1 (strict)
                           A2: Alex is a young adult   (top rule d2, defeasible)
                                 ▲
                                 │ d2 (defeasible)
                           e1: Alex is a student (axiom)

 sub-arguments:  A2 is inside A3, and A3 is inside A4 (not attacks)

Attack graph

        A1 ◀────── rebut ──────▶ A4          A2        A3   (no attacks on them)

Worked solution

State the three attack types
- Rebut: contradictory conclusions, with a defeasible last step.
- Undermine: the conclusion contradicts a defeasible premise.
- Undercut: the conclusion says “rule d does not apply here”.
Identify the conclusions of A1 and A4
- A1 concludes “Alex is not married” via defeasible d1.
- A4 concludes “Alex is married” via defeasible d3 applied to A3. These are contradictory.
Classify the A1-A4 relationship
- Both last steps are defeasible and the conclusions contradict, so it is a rebut.
- Both (A1,A4) and (A4,A1) are rebutting attacks.
Check for other attacks
- A2 (“young adult”) and A3 (“adult”) are contradicted by nobody.
- No argument contradicts a defeasible premise, and none concludes “rule di does not apply”.
- So the only attacks are the mutual rebuts between A1 and A4.

Q6: continue Q5, build AF and evaluate claim

(a) Define AA framework (AR,Att) from Q5’s arguments/attacks. (b) Evaluate “Alex is married” (=conclusion of A4) under admissible vs grounded semantics.

Answer: AR={A1,A2,A3,A4}, Att={(A1,A4),(A4,A1)}. “Alex is married” is admissibly acceptable (A4∈{A4}, an admissible extension) but not groundedly acceptable (GE={A2,A3}, A4∉GE).

Graph

    A2        A3        A1 ◀────▶ A4

Final labelling (grounded, GE = {A2,A3})

   [A2]      [A3]      ?A1? ◀────▶ ?A4?

Worked solution

Construct the framework
- Only attacks: (A1,A4) and (A4,A1). So F = ({A1,A2,A3,A4}, {(A1,A4),(A4,A1)}).
Identify unattacked arguments
- A2 and A3 have no attackers.

Grounded extension by iteration

stage  set        newly defended, and why
f^0    {}         start
f^1    {A2,A3}    A2 and A3 have no attackers
f^2    {A2,A3}    A1 needs an attack on A4, A4 needs an attack on A1: {A2,A3} attacks neither. Fixed point.

GE = {A2,A3}.

Check admissibility of A4

set     conflict-free?  admissible?  reason
{A4}    yes             yes          A1 attacks A4, and A4 attacks A1 back
{A1}    yes             yes          A4 attacks A1, and A1 attacks A4 back

Preferred extensions (script check): {A1,A2,A3} and {A2,A3,A4}.

State the conclusion
- Grounded: A4 ∉ GE, so “Alex is married” is not groundedly acceptable.
- Admissible: A4 ∈ {A4}, so it is admissibly acceptable.
- The claim is accepted under the permissive semantics but not under the most skeptical one (grounded).

Q7: infinite AF, compute GE

AR = {A,D} ∪ {B1,B2,…} ∪ {C1,C2,…}. Att = {(Bn,A),(A,Bn) | n≥1} ∪ {(Cn,Bn) | n≥1} ∪ {(D,Cn) | n≥1}.

Answer: GE = {D}.

Graph (first few nodes, then ”…“)

              D
          ┌───┼───┐   ...           D ──▶ every Cn
          ▼   ▼   ▼
         C1  C2  C3  ...            Cn ──▶ Bn
          │   │   │
          ▼   ▼   ▼
         B1  B2  B3  ...            Bn ◀──▶ A  (two-way, for every n)
          └───┼───┘
              ▼
              A

Final labelling (grounded)

              [D]
          ┌────┼────┐   ...
          ▼    ▼    ▼
        (C1)  (C2)  (C3)  ...          out: attacked by D
          │     │     │
          ▼     ▼     ▼
        ?B1?  ?B2?  ?B3?  ...          undecided
          └─────┼─────┘
                ▼
               ?A?

Worked solution

Find arguments with no attackers
- A is attacked by every Bn. Each Bn is attacked by A and by Cn. Each Cn is attacked by D.
- Nothing attacks D, so D is unattacked.

Grounded extension by stages

stage  set    newly defended, and why
f^0    {}     start
f^1    {D}    only D has no attacker
f^2    {D}    Cn: its only attacker is D and {D} does not attack D, so not defended.
              Bn: attackers are A and Cn; {D} attacks Cn but not A, so not defended.
              A: attacked by every Bn; {D} attacks no Bn, so not defended.

Conclude
- f²=f¹, so the iteration stops after one step: GE = {D}.
- The Bn and A remain undecided: they sit in a symmetric conflict that grounded semantics does not resolve.

Q8: proof, an attacking pair cannot both be groundedly acceptable

“Prove that in an abstract argumentation framework F=(AR,Att), if A,B∈AR and (A,B)∈Att, then it cannot be that both A and B are groundedly acceptable.”

Answer: Proved by contradiction using conflict-freeness of GE.

Proof idea: GE is admissible (Q15), so it is conflict-free. Two members that attack each other would make it conflicted.

   A ──▶ B        if both were in GE, GE would contain an internal attack

Proof

 1. X is groundedly acceptable iff X ∈ GE                       because that is the definition
 2. GE is admissible                                            because of Q15
 3. GE is conflict-free (no X, Y ∈ GE with (X,Y) ∈ Att)         because admissible sets are conflict-free
 4. Assume A ∈ GE and B ∈ GE, with (A,B) ∈ Att                  (for contradiction)
 5. GE contains an internal attack, A attacks B                 because of 4
 6. Contradiction                                               because of 3 and 5
 7. So A and B cannot both be groundedly acceptable.                                           ∎

Q9 (5 marks): describe the steps of argument-based reasoning

“Write a paragraph describing different steps of argument-based reasoning starting from a knowledge base ∆.”

Answer:

Form arguments from ∆, giving a set $A R_{Δ}$ .

Identify attacks between the formed arguments, giving $A t t_{Δ} \subseteq A R_{Δ} \times A R_{Δ}$ .

Form the AA framework $F_{Δ} = (A R_{Δ}, A t t_{Δ})$ .

Select an argumentation semantics $σ$ .

Use $σ$ to evaluate acceptability: $F_{Δ} ⊢_{σ} A$ means $A$ is acceptable under $σ$ .

If $F_{Δ} ⊢_{σ} A$ , extract $A$ ‘s conclusion and treat it as an acceptable claim/inference from $Δ$ .

The pipeline, applied to the Alex example (Q5-Q6 above)

 step                        general                       Alex example
 1 form arguments            AR_Δ from Δ                   A1, A2, A3, A4
 2 identify attacks          Att_Δ                         A1 ◀─rebut─▶ A4
 3 form the framework        (AR_Δ, Att_Δ)                 ({A1..A4}, {(A1,A4),(A4,A1)})
 4 select a semantics σ      grounded / preferred / ...    grounded, or admissible
 5 evaluate acceptability    F ⊢σ A ?                      grounded: GE = {A2,A3}, A4 not in
                                                           admissible: A4 in {A4}
 6 extract conclusions       claim(A)                      "Alex is married" only under admissible

Worked solution

Why start from ∆
- ∆ holds the raw information: facts, strict and defeasible rules, assumptions.
- Arguments are not given directly; they must be constructed by chaining rules from ∆.
Step (1): forming arguments
- Each argument is a finite, acyclic chain of rule applications starting from elements of ∆.
- Sub-arguments are arguments too. The result is the set $A R_{Δ}$ .
Step (2): identifying attacks
- For every pair of arguments, check for a rebutting, undermining or undercutting relationship. This gives $A t t_{Δ}$ .
Step (3): forming the framework
- $(A R_{Δ}, A t t_{Δ})$ is the abstract argumentation framework (AF) for ∆.
- From here on only the attack structure matters, not the content of the arguments.
Steps (4) and (5): semantics and evaluation
- Choose a semantics $σ$ (grounded, preferred, admissible, stable) and compute extensions.
- A is acceptable under $σ$ if it belongs to at least one $σ$ -extension (or to the unique one, for grounded).
Step (6): extraction
- The conclusion of an acceptable argument is a claim that follows from ∆ under the chosen semantics. This is the output.
Connect to Q5-Q6
- ∆ = {d1,d2,s1,d3,e1} gave $A R_{Δ} = {A 1, A 2, A 3, A 4}$ (step 1), and the rebut between A1 and A4 gave $A t t_{Δ}$ (step 2).
- Grounded gave GE = {A2,A3}, so “Alex is married” was not extractable (steps 4-6).
- Admissible gave {A4}, so “Alex is married” is extractable under that weaker semantics.

mt6 (mid term practice 6.pdf)

Q6 (10 marks): infinite “star” framework: admissible, preferred, grounded

F=(AR,Att), AR={A,G}∪{B1,B2,…,Bn,…}, Att={(G,A)}∪{(Bᵢ,G)|i∈ℕ}∪{(Bᵢ,Bⱼ)|i≠j} (every Bᵢ attacks G and every other Bⱼ; G attacks A).

(a) List all admissible extensions. (b) List all preferred extensions. (c) Compute GE.

Answer:

(a) admissible extensions: ∅, {Bᵢ} for each i, {A,Bᵢ} for each i.

(b) preferred extensions: {A,Bᵢ} for each i.

(c) GE = ∅.

Graph (first few nodes, then ”…“)

     B1 ◀────▶ B2 ◀────▶ B3 ...       every pair of B's attacks both ways
      │        │         │
      └────────┴────┬────┘            every Bi attacks G
                    ▼
                    G ───▶ A

Final labelling (preferred extension {A,B1}; the other {A,Bi} are analogous)

     [B1] ◀────▶ (B2) ◀────▶ (B3) ...
       │
       └──────────▶ (G) ───▶ [A]

G is out (B1 attacks it), so its attack on A no longer counts.

Worked solution

Map the attack structure
- G attacks A. Every Bᵢ attacks G and every other Bⱼ. A attacks nobody.
- A is attacked only by G; G by every Bᵢ; each Bᵢ by every Bⱼ with j ≠ i.
Find the conflict-free sets
- ∅ and any singleton {A}, {G}, {Bᵢ} (no self-attacks).
- {A,G} is not (G attacks A). {Bᵢ,Bⱼ} is not (mutual attack). {Bᵢ,G} is not (Bᵢ attacks G).
- {A,Bᵢ} is (neither attacks the other). No set with two different Bᵢ is conflict-free.

Test each set for admissibility

set        conflict-free?  admissible?  reason
{}         yes             yes          nothing to defend
{Bi}       yes             yes          each attacker Bj is attacked by Bi
{A}        yes             no           G attacks A and A attacks nobody
{G}        yes             no           every Bi attacks G; G attacks no Bi
{A,Bi}     yes             yes          Bi attacks G (defends A); Bi attacks every Bj (defends itself)
{A,G}, {Bi,Bj}, {Bi,G}   no  no        contain an attacking pair

Preferred extensions (maximal admissible)
- {Bᵢ} ⊂ {A,Bᵢ}, and {A,Bᵢ} is admissible, so {Bᵢ} is not maximal.
- {A,Bᵢ} cannot grow: another Bⱼ or G would conflict with Bᵢ. So each {A,Bᵢ} is preferred.

Grounded extension

stage  set   newly defended, and why
f^0    {}    start
f^1    {}    A is attacked by G, G by every Bi, every Bi by every other Bj: nobody is unattacked

GE = ∅: infinitely many preferred extensions, yet grounded semantics is maximally skeptical here.

Q7 (10 marks): proof, an argument attacked by GE cannot be admissibly acceptable

“Prove that in any abstract argumentation framework F=(AR,Att), if an argument A∈AR is attacked by some argument in the grounded extension GE, then A cannot be admissibly acceptable.”

Answer: Proved by contradiction, by induction on the layers $f^{i} (\emptyset)$ that build up GE.

Proof idea: an admissible set $E$ containing $A$ would have to attack $B \in GE$ . But $E$ can never attack any stage of GE: an attack on stage $k + 1$ is answered from stage $k$ , and $E$ would then have to counter-attack that answer, which lies in stage $k$ .

Picture of the contradiction

   B ──▶ A          B in GE, A in E
   ▲
   C                C in E must attack B (E defends A), yet E attacks nothing in GE

Proof

Lemma R(k): E attacks no member of f^k(∅), where E is any admissible set
 1. R(0) holds                                    because f^0(∅) = ∅ has no members
 2. Assume R(k). Suppose D ∈ E attacks X ∈ f^(k+1)(∅)     (for contradiction)
 3. Some Y ∈ f^k(∅) attacks D                     because f^k(∅) defends X and D attacks X
 4. E attacks Y                                   because E is admissible, D ∈ E, and Y attacks D
 5. E attacks a member of f^k(∅)                  because Y ∈ f^k(∅); this contradicts R(k)
 6. R(k+1) holds
 7. E attacks no member of GE                     because GE is the union of the stages f^k(∅)

Main claim
 8. Suppose B ∈ GE attacks A, and A ∈ E for an admissible E      (for contradiction)
 9. Some C ∈ E attacks B                          because E defends A and B attacks A
10. E attacks a member of GE                      because B ∈ GE
11. Contradiction                                 because of 7
12. So A is in no admissible extension: A is not admissibly acceptable.                    ∎

Bonus Q: stable extensions

Worked examples of an AF with no stable extension (a single self-attacker, e.g. AR={A}, Att={(A,A)}) and one with multiple stable extensions (mutual 2-cycle, Att={(A,B),(B,A)}, giving stable extensions {A} and {B}).

Answer: self-attacker case: no stable extension exists. 2-cycle case: {A} and {B} are both stable extensions.

Graphs

   self-attacker              2-cycle
       A ↺                    A ◀────▶ B

Final labellings

   self-attacker: no stable set     2-cycle: stable {A}    stable {B}
       ?A? ↺                                 [A] ◀▶ (B)     (A) ◀▶ [B]

Worked solution

Recall the definition
- S is stable if (i) it is conflict-free and (ii) it attacks every argument outside S.

Self-attacker case (AR={A}, Att={(A,A)})

set    conflict-free?  attacks everything outside?  stable?  reason
{}     yes             no                            no       {} attacks nothing, so A stays unattacked
{A}    no              (not needed)                  no       A attacks itself

Neither subset satisfies both conditions: no stable extension.

2-cycle case (AR={A,B}, Att={(A,B),(B,A)})

set    conflict-free?  attacks everything outside?  stable?  reason
{A}    yes             yes: A attacks B              yes
{B}    yes             yes: B attacks A              yes
{}     yes             no                            no       attacks nothing
{A,B}  no              (not needed)                  no       mutual attack

Contrast with the grounded extension
- The grounded extension is always unique.
- Stable extensions can fail to exist (self-attacker) or be multiple (2-cycle).
- Every stable extension is also preferred, but not every preferred extension is stable.

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practice_extracted

Practice Questions: Arguments & Propositions in NL, and Computational Argumentation

Part 1: Arguments & Propositions in Natural Language

mt1 (mid term practice 1.pdf)

Q1 (4 marks): Fill in the blanks

Q2 (4 marks): Mark reason (R) and conclusion (C)

Q3 (4 marks): Extract main conclusion from editorial text

Q4 (4 marks): Which passages are arguments; identify conclusion

Q5 (4 marks): Deductive or inductive?

Q11 (5 marks): Best categorical form (A/E/I/O) for each NL proposition

mt2 (mid term practice 2.pdf)

Q1: Is it an argument?

Q2: Extract main conclusion from editorial

Q3: Syllogism validity by Venn diagram

Q4: Syllogism validity

Q5: Syllogism validity

Q6: Syllogism validity

mt3 (mid term practice 3.pdf)

Q1: Identify argument + standard form

Q2: Suppressed premises, standard form

Q3: Extract 2+ arguments from text, using argument-map hints

Q4: Syllogism validity

Q5: Syllogism validity

Q6: Syllogism validity

Q7: Syllogism validity

mt5 (mid term practice 5.pdf)

Q1 (3 marks): Which sentence is an argument?

Q2 (3 marks): Which sentence is NOT an argument?

Q3 (3 marks): Odd one out

Q4 (3 marks): Which argument is deductive?

Q5 (3 marks): Which sentence cannot be a conjunction of two simpler sentences?

Q6, Q7, Q8: Which Venn diagram correctly represents the given categorical proposition(s)?

mt6 (mid term practice 6.pdf)

Q1 (10 marks): Does the proposition have the form ¬S?

Q2 (10 marks): Best categorical form for NL propositions

Q3 (10 marks): Suppressed premises for deductive validity

Q4 (10 marks): Form arguments and identify attacks

Q5 (10 marks): Barber paradox arguments

Part 2: Computational Argumentation

How to read the diagrams

mt5 (mid term practice 5.pdf)

Q9 (5 marks): full mutual 3-cycle

Q10 (5 marks): directed 3-cycle

Q11

Q12: infinite backward chain

Q13 / Q14: ASPIC+ quarantine(bob)

Q15: GE is admissible (proof)

mt4 (mid term practice 4.pdf, answer-key excerpt)

Q17: two arguments A, B (statement truncated in source)

Q18: AR={A0,A1,A2}, Att={(A1,A0),(A2,A1),(A0,A2),(A2,A2)}

Q19: AR={A0,A1,A2,A3}, Att={(A1,A0),(A2,A1),(A0,A2),(A3,A2)}

Q20: AR={A0,A1,A2}, Att={(A1,A0),(A2,A0),(A1,A2),(A2,A1)} (“double cycle”)

mt7 (mid term practice 7.pdf)

Q5: knowledge base to arguments

Q6: continue Q5, build AF and evaluate claim

Q7: infinite AF, compute GE

Q8: proof, an attacking pair cannot both be groundedly acceptable

Q9 (5 marks): describe the steps of argument-based reasoning

mt6 (mid term practice 6.pdf)

Q6 (10 marks): infinite “star” framework: admissible, preferred, grounded

Q7 (10 marks): proof, an argument attacked by GE cannot be admissibly acceptable

Bonus Q: stable extensions

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