Question 1 of 12

Question 1

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Four independent predicate-logic statements built from the standard quantifier negation laws, an implication of two nested quantifiers, and Fermat's Last Theorem as a target sentence.

Find. (a) an equivalent quantifier-free-of-negation form; (b) the symbolic translation; (c) whether the implication is logically valid, with proof or counterexample; (d) a fully quantified logical sentence encoding Fermat's Last Theorem.

Approach. Push every negation through a quantifier one step at a time using ¬∀x Q(x) ≡ ∃x ¬Q(x) and ¬∃x Q(x) ≡ ∀x ¬Q(x); translate "necessary condition" as the forward direction of an implication; test the boxed implication in (c) with a small finite model before attempting a proof; and assemble (d) from an outer ∀n over the exponent domain wrapped around a negated existential over the three positive-integer witnesses.

1. Part (a) — strip the negation inward. $$\lnot\exists x\,\forall y\,\exists z\;P(x,y,z)$$ Apply ¬∃ → ∀¬, then ¬∀ → ∃¬, then ¬∃ → ∀¬ once more, moving one quantifier at a time: $$\lnot\exists x\,\forall y\,\exists z\,P \equiv \forall x\,\lnot\forall y\,\exists z\,P \equiv \forall x\,\exists y\,\lnot\exists z\,P \equiv \boxed{\forall x\,\exists y\,\forall z\;\lnot P(x,y,z)}$$ Every quantifier flips type (∀↔∃) and the negation lands on the innermost matrix only.
2. Part (b) — "necessary condition" as an implication. "A necessary condition for P(x,y) is that x>y" means P(x,y) being true forces x>y to be true — P(x,y) cannot hold unless x>y already holds. That is exactly a conditional with P(x,y) as antecedent: $$\boxed{P(x,y) \rightarrow x>y}$$ (x>y is necessary for P but not claimed sufficient, so no biconditional.)
3. Part (c) — test ∃x∀yP(x,y) → ∀x∃yP(x,y) for validity. Try the smallest possible countermodel: domain D = {1, 2}, and define $$P(x,y) \equiv (x = 1)$$ Evaluate the antecedent: taking x = 1 gives P(1,y) true for every y, so ∃x∀yP(x,y) is true. Evaluate the consequent: for x = 2, P(2,y) is false for every y (since 2≠1), so no y makes P(2,y) true and ∀x∃yP(x,y) is false. An implication with a true antecedent and a false consequent is false, so the statement fails on this model. $$\boxed{\text{NOT a tautology --- countermodel: } D=\{1,2\},\ P(x,y)\equiv(x=1)}$$ The reused variable name "x" on both sides is the trap: the witness x that makes ∀y true on the left is not obliged to be the same x that must supply a y on the right for every x, including ones that behave nothing like the witness.
4. Part (d) — Fermat's Last Theorem as a sentence. Let the universe of discourse be the positive integers throughout, and let Pⁿ(x,y,z) ≡ (xⁿ+yⁿ=zⁿ). The claim "for every exponent n>2 there is no positive-integer solution" is $$\boxed{\forall n\;\big[(n>2)\rightarrow \lnot\exists x\,\exists y\,\exists z\;P_n(x,y,z)\big]}$$ equivalently, pushing the negation inward, ∀n[(n>2)→∀x∀y∀z\,¬Pⁿ(x,y,z)].