Which one of the following is NOT logically equivalent to \(¬∃x (∀ y (α)∧∀z(β…

Derivation of the given formula: ¬∃x (∀y α ∧ ∀z β)

• Apply ¬∃x P ≡ ∀x ¬P: ∀x ¬(∀y α ∧ ∀z β).

• Use De Morgan: ∀x (¬∀y α ∨ ¬∀z β).

• Convert ¬∀ to ∃¬: ∀x (∃y ¬α ∨ ∃z ¬β).

Compare each candidate formula to the derived form ∀x (∃y ¬α ∨ ∃z ¬β):

1. Candidate: ∀x (∃z ¬β → ∀y α).

   Rewrite: (∃z ¬β → ∀y α) ≡ (¬∃z ¬β ∨ ∀y α) ≡ (∀z β ∨ ∀y α).

   Verdict: Not equivalent. Example: if for some x both α holds for all y and β holds for all z, then the derived form is false for that x but the candidate is true.

2. Candidate: ∀x (∀z β → ∃y ¬α).

   Rewrite: (∀z β → ∃y ¬α) ≡ (¬∀z β ∨ ∃y ¬α) ≡ (∃z ¬β ∨ ∃y ¬α).

   Verdict: Equivalent. This matches ∀x (∃y ¬α ∨ ∃z ¬β) up to the order of disjuncts.

3. Candidate: ∀x (∀y α → ∃z ¬β).

   Rewrite: (∀y α → ∃z ¬β) ≡ (¬∀y α ∨ ∃z ¬β) ≡ (∃y ¬α ∨ ∃z ¬β).

   Verdict: Equivalent. This matches the derived form exactly.

4. Candidate: ∀x (∃y ¬α → ∃z ¬β).

   Rewrite: (∃y ¬α → ∃z ¬β) ≡ (¬∃y ¬α ∨ ∃z ¬β) ≡ (∀y α ∨ ∃z ¬β).

   Verdict: Not equivalent. Example: for some x let α be false for some y (so ∃y ¬α is true) while β holds for all z (so ∃z ¬β is false). The derived form is true for that x but this candidate is false.

Conclusion: The second and third candidate formulas are equivalent to the original formula after standard transformations. The first and fourth candidate formulas are not equivalent; they fail in simple counterexamples as shown. Therefore the answer key that identifies only the first candidate as not equivalent is incomplete.