practice_gen_comp

Practice Questions (Generated): Computational Argumentation

These are newly generated questions (not from the original exam bank in practice_extracted.md). They cover every question style in that bank for this topic:

• finite AF graph analysis
• infinite AF analysis
• ASPIC+ construction (including an undercutting attack, which the original bank never computed)
• self-referential paradox arguments
• formal proofs

Every computation was checked by a small script (conflict-free, admissible, preferred, stable, grounded) and every fixed-point iteration is shown as a table so you can re-derive it.

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

Grounded extension recipe: start from the empty set, keep adding every argument that the current set defends, stop when nothing new is added.

---

Part A: Finite argumentation frameworks

Q1 (multiple choice): the “diamond with a twist”

$F=(AR,Att)$, $AR=\{A,B,C,D\}$, $Att=\{(A,B),(B,C),(A,D),(D,C),(C,A)\}$ ($A$ attacks $B$ and $D$; $B$ attacks $C$; $D$ attacks $C$; $C$ attacks $A$).

How many arguments are admissibly acceptable? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4

Answer: (a) 0

Graph (the diagonal is $C⟶A$)

    A ───────▶ B
    │ ◀╲       │
    │    ╲     │
    ▼      ╲   ▼
    D ───────▶ C

Final labelling: no argument is in, none is out, all are undecided.

    ?A? ─────▶ ?B?
     │ ◀╲        │
     ▼    ╲      ▼
    ?D? ─────▶ ?C?

Worked solution

1. List who attacks whom

   • A is attacked by C. B is attacked by A. C is attacked by B and D. D is attacked by A.
   • Every argument has an attacker, so nobody is “free”.

2. Find all conflict-free sets

   • Five pairs conflict: $\{A,B\}$, $\{A,C\}$, $\{A,D\}$, $\{B,C\}$, $\{C,D\}$.
   • Only $\{B,D\}$ survives: neither $(B,D)$ nor $(D,B)$ is in $Att$.
   • Any triple contains a conflicting pair, so none is conflict-free.

3. Test each set for admissibility

   set      conflict-free?  admissible?  reason
   {}       yes             yes          nothing to defend
   {A}      yes             no           C attacks A; A attacks B, D but not C
   {B}      yes             no           A attacks B; B attacks only C
   {C}      yes             no           B, D attack C; C attacks only A
   {D}      yes             no           A attacks D; D attacks only C
   {B,D}    yes             no           A attacks B and D; neither B nor D attacks A
   others   no              no           contain an attacking pair
   

4. Grounded extension by iteration

   stage  set   newly defended, and why
   f^0    {}    start
   f^1    {}    no argument is unattacked, so nothing is defended by {}
   f^2    {}    fixed point, GE = {}
   

5. Conclude

   • The only admissible extension is $\varnothing$, so zero arguments are admissibly acceptable: (a).
   • Side notes: no stable extension exists ($\{B,D\}$ does not attack $A$), and $GE=\varnothing$.

---

Q2 (open): self-attack plus a branch

$AR=\{A_0,A_1,A_2,A_3\}$, $Att=\{(A_0,A_0),(A_0,A_1),(A_2,A_1),(A_2,A_3)\}$ ($A_0$ self-attacks and attacks $A_1$; $A_2$ attacks both $A_1$ and $A_3$).

Find all conflict-free sets, all admissible extensions, the preferred extension(s), and $GE$.

Answer:

• conflict-free $=\{\varnothing ,\{A_1\},\{A_2\},\{A_3\},\{A_1,A_3\}\}$
• admissible $=\{\varnothing ,\{A_2\}\}$
• preferred $=\{\{A_2\}\}$
• $GE=\{A_2\}$

Graph

    A0 ↺ ───▶ A1 ◀─── A2 ───▶ A3

Final labelling (extension $\{A_2\}$)

    ?A0? ↺ ───▶ (A1) ◀─── [A2] ───▶ (A3)

$A_0$ stays undecided: it attacks itself, and nothing in the extension attacks it.

Worked solution

1. Handle the self-attack first

   • $(A_0,A_0)\in Att$, so no conflict-free set contains $A_0$.
   • $A_0$ still attacks $A_1$ from outside.

2. Find conflict-free sets among $\{A_1,A_2,A_3\}$

   • $A_2$ attacks $A_1$ and $A_3$, so $\{A_1,A_2\}$ and $\{A_2,A_3\}$ conflict.
   • $\{A_1,A_3\}$ has no attack either way, so it is conflict-free.

3. Test each set for admissibility

   set        conflict-free?  admissible?  reason
   {}         yes             yes          nothing to defend
   {A1}       yes             no           attacked by A0 and A2; A1 attacks neither
   {A2}       yes             yes          unattacked, so vacuously defended
   {A3}       yes             no           attacked by A2; A3 attacks nobody
   {A1,A3}    yes             no           A0 attacks A1 and nothing in the set attacks A0
   with A0    no              no           self-attack
   others     no              no           contain an attacking pair
   

4. Find the preferred extension

   • $\{A_2\}$ is admissible. Adding $A_1$ or $A_3$ creates a conflict with $A_2$.
   • So $\{A_2\}$ is maximal: the unique preferred extension.

5. Grounded extension by iteration

   stage  set    newly defended, and why
   f^0    {}     start
   f^1    {A2}   A2 has no attacker
   f^2    {A2}   A1 needs attacks on A0 and A2: none. A3 needs an attack on A2: none.
   

   • $f^2=f^1$, so $GE=\{A_2\}$. Admissible, preferred and grounded coincide.
   • The self-attacking $A_0$ still blocks $A_1$, like the “poisoned chain” case in the theory file, even though $A_0$ is in no conflict-free set.

---

Q3 (multiple choice): a shared victim

$AR=\{A,B,C\}$, $Att=\{(A,B),(B,A),(A,C),(B,C)\}$ ($A$ and $B$ attack each other; both also attack $C$).

Which of the following is false? (a) $\{A\}$ and $\{B\}$ are both admissible. (b) $C$ is admissibly acceptable under at least one admissible extension. (c) $GE=\varnothing$. (d) There are exactly two preferred extensions. (e) $C$ is never in any admissible extension.

Answer: (b) is false.

Graph

        A ◀──────▶ B
         ╲         ╱
          ╲       ╱
           ▼     ▼
              C

Final labellings

   preferred {A}        preferred {B}        grounded (GE = {})
   [A] ◀──▶ (B)         (A) ◀──▶ [B]         ?A? ◀──▶ ?B?
     ╲      ╱             ╲      ╱             ╲      ╱
      ▼    ▼               ▼    ▼               ▼    ▼
        (C)                  (C)                  ?C?

Worked solution

1. Find conflict-free sets

   • $\{A,B\}$, $\{A,C\}$, $\{B,C\}$ all contain an attack.
   • Only $\varnothing ,\{A\},\{B\},\{C\}$ are conflict-free.

2. Test each set for admissibility

   set    conflict-free?  admissible?  reason
   {}     yes             yes          nothing to defend
   {A}    yes             yes          B attacks A, but A attacks B back
   {B}    yes             yes          A attacks B, but B attacks A back
   {C}    yes             no           A and B attack C; C attacks neither
   

3. Check the options

   • (a) true: shown in the table.
   • (e) true: $C$ is in no admissible set, and this shows (b) is false.
   • $C$ is a bystander: both warring sides attack it and it cannot hit back.

4. Preferred extensions

   • $\{A\}$ and $\{B\}$ are each maximal: adding the other conflicts, adding $C$ conflicts.
   • Exactly two preferred extensions: (d) true.

5. Grounded extension

   stage  set   newly defended, and why
   f^0    {}    start
   f^1    {}    A is attacked by B, B by A, C by both: nobody is unattacked
   

   • $GE=\varnothing$: (c) true.

6. Conclude

   • Every statement except (b) is true, so (b) is the false one.

---

Part B: Infinite argumentation frameworks

Q4 (open): an infinite fan without infighting

$AR=\{A\}\cup \{B_1,B_2,B_3,\dots \}$, $Att=\{(B_i,A)\mid i\ge 1\}$ only. Every $B_i$ attacks $A$; the $B_i$‘s do not attack each other, unlike the “star” framework from the original bank where they do.

List the admissible extensions, the preferred extension(s), and compute $GE$.

Answer:

• admissible $=$ every subset of $\{B_1,B_2,\dots \}$ (including $\varnothing$ and the full set)
• preferred $=$ the single set $\{B_1,B_2,\dots \}$ (all $B_i$‘s, excluding $A$)
• $GE=\{B_1,B_2,\dots \}$

Graph (first few nodes, then ”…“)

     B1     B2     B3     B4    ...      (no arrows between the B's)
      │      │      │      │
      └──────┴───┬──┴──────┘
                 ▼
                 A

Final labelling (extension = all the $B_i$)

     [B1]   [B2]   [B3]   [B4]   ...
       └──────┴───┬──┴──────┘
                  ▼
                 (A)

Worked solution

1. No $B_i$ has an attacker

   • $Att$ only has pairs $(B_i,A)$, so nothing attacks a $B_i$.
   • Any subset of the $B_i$‘s is conflict-free (they never attack each other) and vacuously admissible (no attackers to defend against).
   • There is one such subset for every subset of an infinite set: uncountably many admissible extensions.

2. $A$ can never be admissible

   • $A$‘s attackers are all the $B_i$‘s, and $A$ attacks nobody.
   • So $\{A\}$ is undefended and not admissible.
   • Any set containing $A$ and some $B_i$ is not conflict-free.

3. The preferred extension

   • Adding a missing $B_j$ to any proper subset keeps it conflict-free and admissible.
   • So the only maximal admissible set is the full set $\{B_1,B_2,\dots \}$: the unique preferred extension.

4. Grounded extension by stages

   stage  set                newly defended, and why
   f^0    {}                 start
   f^1    {B1, B2, B3, ...}  every B_i is unattacked: all join at once
   f^2    {B1, B2, B3, ...}  A would need attacks on every B_i: nothing attacks any B_i
   

   • $GE=\{B_1,B_2,\dots \}$.
   • The whole infinite set arrives in one application of $f$, unlike the chain examples in the theory file where $GE$ grows one argument per stage. Each $B_i$ is independently unattacked, so there is no chain.

5. Compare with the original bank’s “star” framework

   • Star: $AR=\{A,G\}\cup \{B_i\}$, $Att=\{(G,A)\}\cup \{(B_i,G)\}\cup \{(B_i,B_j)\mid i\ne j\}$.

                        this question        star framework
   B_i attack each other  no                    yes
   what attacks A         every B_i             only G (and every B_i attacks G)
   preferred              one set: all B_i      {A,B_i}, one for each i
   GE                     all B_i               empty
   

   • In the star, the $B_i$‘s compete, so at most one of them is in an extension. They also all attack $G$, which frees $A$, giving one preferred extension $\{A,B_i\}$ per $i$.
   • Lesson: whether the many attackers of a shared target also fight each other completely changes the extension count.

Note (check): the original said adding $(B_i,B_j)$ attacks here gives preferred $\{A,B_i\}$. A script shows it gives $\{B_i\}$ (A stays out, GE empty). The $\{A,B_i\}$ sets come from the star’s extra argument $G$.

---

Part C: ASPIC+ structured argumentation, including an undercutting attack

Q5 (10 marks): express-delivery eligibility

Knowledge base $\Delta$ for whether package $P$ is eligible for express delivery:

• $d_1$: “Packages under 2kg are normally express-eligible.”
• $e_1$: “Package $P$ weighs 1.5kg.” (axiom)
• $d_2$: “International packages are normally not express-eligible.”
• $e_2$: “Package $P$ is international.” (axiom)
• $d_3$: “A package with a verified pre-clearance customs stamp means rule $d_2$ does not apply.”
• $e_3$: “Package $P$ has a verified pre-clearance customs stamp.” (axiom)

(a) Form the arguments $A_1=\{e_1,d_1\}⊢$ ”$P$ is express-eligible”, $A_2=\{e_2,d_2\}⊢$ ”$P$ is not express-eligible”, $A_3=\{e_3,d_3\}⊢$ ”$d_2$ does not apply”. Identify all attacks. (b) Build the framework and evaluate ”$P$ is express-eligible” under admissible and grounded semantics.

Answer:

• (a) $A_1$ and $A_2$ rebut each other; $A_3$ undercuts $A_2$ (it attacks the applicability of $d_2$, not its conclusion). $Att=\{(A_1,A_2),(A_2,A_1),(A_3,A_2)\}$.
• (b) $GE=\{A_1,A_3\}$: ”$P$ is express-eligible” is groundedly acceptable, because the undercutter breaks the tie in $A_1$‘s favour.

The three arguments (premises at the leaves, conclusion at the root)

 A1                          A2                            A3
 P is express-eligible       P is not express-eligible     d2 does not apply
        ▲                           ▲                             ▲
        │ d1 (defeasible)           │ d2 (defeasible)             │ d3 (defeasible)
 e1: P weighs 1.5kg          e2: P is international        e3: P has a pre-clearance stamp
     (axiom)                     (axiom)                       (axiom)

Attack graph

   A1 ◀──── rebut ────▶ A2 ◀──── undercut ──── A3

Final labelling

   [A1] ◀──── rebut ────▶ (A2) ◀──── undercut ──── [A3]

Worked solution

1. Classify $A_1$ versus $A_2$

   • $A_1$ concludes “express-eligible”, $A_2$ concludes “not express-eligible”: directly contradictory.
   • Both last steps are defeasible ($d_1$, $d_2$), so it is a rebut in both directions: $(A_1,A_2)$ and $(A_2,A_1)$.

2. Classify $A_3$ versus $A_2$

   • $A_3$ concludes “rule $d_2$ does not apply”, and $A_2$‘s last step used exactly $d_2$.
   • That is an undercut: it targets the licence of $A_2$‘s inference, not its conclusion or premises. So $(A_3,A_2)\in Att$.
   • It goes one way only: $A_2$‘s conclusion says nothing about $d_3$.

3. Build the framework

   • $AR=\{A_1,A_2,A_3\}$, $Att=\{(A_1,A_2),(A_2,A_1),(A_3,A_2)\}$.

4. 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 only attacker is A2, and A3 attacks A2, so A1 is defended.
                     A2 is not: its attackers are A1 and A3, and nothing in {A3} attacks A1.
   f^3    {A1,A3}    A2 would need attacks on both A1 and A3: none. Fixed point.
   

5. Check admissible, preferred, stable

   set        conflict-free?  admissible?  reason
   {A1}       yes             yes          A2 attacks A1, A1 attacks A2 back
   {A2}       yes             no           A3 attacks A2 and A2 cannot answer A3
   {A3}       yes             yes          unattacked
   {A1,A3}    yes             yes          A1 defended by A3; preferred, and stable (A1 attacks A2)
   {A1,A2}    no              no           rebut
   {A2,A3}    no              no           undercut
   

   • $\{A_1,A_3\}$ is the unique preferred extension and is also stable.
   • So admissible, preferred and grounded semantics all agree: express-eligible.

6. Interpret

   • $A_3$ is unattacked, so it is unconditionally accepted, and it knocks $A_2$ out. This resolves the symmetric rebut in $A_1$‘s favour.
   • A bare symmetric rebut (like the original bank’s Alex-is-married example) leaves both sides out of $GE$.
   • Undercut can break a tie that rebut alone cannot.

---

Part D: A self-referential paradox (same structure as the barber paradox)

Q6 (10 marks): the crocodile’s promise

A crocodile steals a child and tells the father: “I will return your child if and only if you correctly guess whether I will return the child.” The father replies: “You will not return my child.”

Construct the two arguments that follow from assuming each outcome, identify all attacks, and explain what the framework reveals.

Answer:

• $A$ = assume “returns”, derive “does not return” (self-contradicting).
• $B$ = assume “does not return”, derive “returns” (self-contradicting).
• $A$ and $B$ rebut each other; each also undermines itself: $Att=\{(A,B),(B,A),(A,A),(B,B)\}$.
• No admissible extension other than $\varnothing$ exists, and there is no stable extension.

The two arguments

 A                                           B
 does not return                             returns
      ▲                                           ▲
      │ rule R: returns iff guess is correct      │ rule R
 guess "will not return" is incorrect        guess "will not return" is correct
      ▲                                           ▲
      │                                           │
 assume: returns  (defeasible premise)       assume: does not return  (defeasible premise)

Attack graph (the four attacks)

     ↺                        ↺
     A ◀────── rebut ───────▶ B
  (undermines itself)      (undermines itself)

Final labelling: everything is undecided (no argument is in or out).

     ↺                        ↺
    ?A? ◀────────────────────▶ ?B?

Worked solution

1. Formalise argument $A$

   • Rule: the crocodile returns the child if and only if the father’s guess is correct.
   • Assume it returns. The father guessed “will not return”, so the guess is incorrect.
   • By the rule, an incorrect guess means it does not return: this contradicts the assumption.

2. Formalise argument $B$

   • Same rule. Assume it does not return. The guess “will not return” is correct.
   • By the rule, a correct guess means it does return: again a contradiction with the assumption.

3. Identify the rebut

   • $A$ concludes “does not return”, $B$ concludes “returns”: direct contradiction, so $(A,B)$ and $(B,A)$.

4. Identify the self-undermining

   • $A$‘s conclusion contradicts $A$‘s own premise (the assumption “returns”): $(A,A)$. By symmetry, $(B,B)$.

Note (check): $B$‘s conclusion also contradicts $A$‘s assumption, so $(B,A)$ can be read as an undermine as well as a rebut. The four-arrow framework is the same either way.

5. Assemble the framework and check admissibility

   set      conflict-free?  admissible?  stable?  reason
   {}       yes             yes          no       attacks nothing, so A and B stay outside unattacked
   {A}      no              no           no       A attacks itself
   {B}      no              no           no       B attacks itself
   {A,B}    no              no           no       self-attacks and mutual attack
   

   • Grounded: $f^1=\varnothing$ because $A$ is attacked by $B$ and itself, and $B$ by $A$ and itself. $GE=\varnothing$.

6. The point

   • Any self-referential “X if and only if not-X” promise gives this same four-attack signature (as in the barber paradox): each assumption undermines itself once its conclusion is derived, and the pair also rebuts.
   • The collapse to $\varnothing$ mirrors the informal fact that the promise is logically unsatisfiable: no consistent outcome exists.

---

Part E: Proofs

Q7 (10 marks): the grounded extension is a subset of every stable extension

“Let $F=(AR,Att)$ be an argumentation framework, and suppose $S$ is a stable extension of $F$ (i.e. $S$ is conflict-free and attacks every argument outside $S$). Prove that $GE\subseteq S$.”

Answer: Proved by induction on the stages $f^i(\varnothing )$ that build up $GE$, using totality of stable extensions (every argument is either in $S$ or attacked by $S$).

Proof idea: if $S$ defends $X$ but $X\notin S$, then some $Z\in S$ attacks $X$, and $S$ must attack $Z$ from inside: an attack inside $S$, contradicting conflict-freeness.

Picture of the contradiction

   Z ──▶ X        Z is in S (totality: X is outside S, so S attacks X)
   ▲
   W              W is in S (S defends X, so S attacks every attacker of X, including Z)
                  W and Z both in S with W ──▶ Z: S is not conflict-free

Proof

Facts used
 (i)  Defense is monotone: T ⊆ T' and T defends X  imply  T' defends X.
 (ii) Totality: for every X, either X ∈ S or some member of S attacks X.

Claim P(i):  f^i(∅) ⊆ S,   for all i ≥ 0

 1. P(0) holds                                    because f^0(∅) = ∅ ⊆ S
 2. Assume P(i). Take X ∈ f^(i+1)(∅)              because f^(i+1)(∅) = f(f^i(∅))
 3. f^i(∅) defends X                              because X ∈ f(f^i(∅))
 4. S defends X                                   because of 3, P(i) and fact (i)
 5. Suppose X ∉ S                                 (for contradiction)
 6. Some Z ∈ S attacks X                          because of 5 and fact (ii)
 7. S attacks Z, say W ∈ S with W ──▶ Z           because of 4 and 6 (S defends X)
 8. W, Z ∈ S and W attacks Z                      because of 6 and 7
 9. Contradiction                                 because S is conflict-free
10. So X ∈ S, hence P(i+1)                        because of 5 to 9
11. GE = ⋃ f^i(∅) ⊆ S                             because P(i) holds for every i
                                                                        ∎

Why it matters

• A stable extension is always a preferred extension, so this also places $GE$ inside every stable one: grounded semantics is at least as skeptical as any stable extension.
• Example: in Q3, $GE=\varnothing$ is a subset of the stable extensions $\{A\}$ and $\{B\}$.

---

Q8 (5 marks): true or false, with justification

For an arbitrary argumentation framework $F=(AR,Att)$, is the following statement true or false: “If argument $X$ is admissibly acceptable, then $X$ is groundedly acceptable.”

Answer: False.

Counterexample: the Q3 framework

   A ◀──▶ B                       admissible sets:  {}  {A}  {B}
    ╲    ╱                        GE:               {}
     ▼  ▼
      C

   A is in the admissible set {A}, but A is not in GE = {}.

Worked solution

1. Recall the definitions

   • $X$ is admissibly acceptable if $X$ is in some admissible extension.
   • $X$ is groundedly acceptable if $X\in GE$, the unique grounded extension.

2. Relate the two notions

   • $GE$ is itself an admissible extension, so groundedly acceptable implies admissibly acceptable.
   • The statement claims the reverse direction.

Note (check): the original said $GE$ is “the least” admissible extension. That is not right: $\varnothing$ is always admissible and is contained in $GE$. $GE$ is the least complete extension. The conclusion here is unaffected.

3. Produce the counterexample

   • Take Q3: $AR=\{A,B,C\}$, $Att=\{(A,B),(B,A),(A,C),(B,C)\}$.
   • $A$ is in the admissible extension $\{A\}$, but $GE=\varnothing$, so $A\notin GE$.

4. Conclude

   • The statement is false: admissible acceptability is strictly more permissive than grounded acceptability in general.
   • They coincide in special cases, e.g. when a unique preferred extension equals $GE$ (as in Q2, Q4, Q5 above).
   • They differ with several competing preferred extensions (Q3, or the original bank’s mutual-cycle and backward-chain examples).