Which one of the following is a theorem in the Propositional Calculus?

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Question: Which one of the following is a theorem in the Propositional Calculus?

Option given: ∼(P ∧ Q) ⇒ (∼P ∨ ∼Q)

First, set up what the statement is saying. It asserts that the implication from ¬(P ∧ Q) to (¬P ∨ ¬Q) holds universally for all truth values of P and Q. In other words, is the formula ¬(P ∧ Q) → (¬P ∨ ¬Q) a tautology?

A helpful angle is to compare the antecedent with the consequent. By De Morgan’s law, ¬(P ∧ Q) is logically equivalent to (¬P ∨ ¬Q). This equivalence means that whenever ¬(P ∧ Q) is true, ¬P ∨ ¬Q is also true, and in fact they are true in exactly the same circumstances.

Given that the antecedent and the consequent are equivalent, the implication ¬(P ∧ Q) → (¬P ∨ ¬Q) is true in every possible truth assignment for P and Q. In formal logic terms, a statement of the form A → B is a tautology if A and B are equivalent (A ↔ B is a tautology). Here A is ¬(P ∧ Q) and B is (¬P ∨ ¬Q), and A ↔ B is indeed a tautology by De Morgan’s law.

Now, consider what this tells us about the nature of the given formula. Since De Morgan’s law directly provides the equivalence between ¬(P ∧ Q) and (¬P ∨ ¬Q), the implication is always true for all truth values of P and Q. Therefore, the formula is a theorem of propositional calculus because it is a valid (tautological) instance.

Note on the structure of the question: only one option is presented, and there are no other choices to compare against. If there were multiple options, we would similarly assess each candidate for tautologicity or derivability within standard propositional axioms.

Which one of the following is a theorem in the Propositional Calculus?

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