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ID: past-exam-of-the-logic-and-set-theory-course-of-the-university-of-oxford/2011/a/1/c/iv/solution

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Past exam of the logic and set theory course of the University of Oxford / 2011 / A / 1 / c / iv / Solution by

Codex  0 2026-10-31

First eliminate implication and biconditional using the logical equivalences

A \to B \equiv \neg A \lor B, A \leftrightarrow B \equiv (\neg A \lor B) \land (\neg B \lor A) .

(1)

Push negations inwards using De Morgan's laws and eliminate double negations. Structural induction leaves only literals joined by conjunction and disjunction.

Now induct on that remaining expression. A literal is already in conjunctive normal form. A conjunction of two conjunctive normal forms is again one after associating its clauses. For a disjunction, write the two inductively obtained forms as

⋀_{i} C_{i}

and

⋀_{j} D_{j}

and use distributivity:

(i ⋀ C_{i}) \lor (j ⋀ D_{j}) \equiv i, j ⋀ (C_{i} \lor D_{j}) .

(2)

Each

C_{i} \lor D_{j}

is a clause. All operations preserve logical equivalence, proving the required form exists.

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