Which of the following are tautology? (A) (P → (P ∧ Q)) → (P → Q) (B) ((P → Q)…

Final answer: the first and second propositions are tautologies; the third and fourth are not.

• First proposition: (P → (P ∧ Q)) → (P → Q). The subformula P → (P ∧ Q) is logically equivalent to P → Q (since P → (P ∧ Q) ≡ ¬P ∨ Q), so the implication is of the form A → A and is always true.

• Second proposition: ((P → Q) → Q) → (P ∨ Q). By cases: if Q is true then P ∨ Q is true; if Q is false then (P → Q) → Q becomes (P → false) → false, which simplifies to P, and P implies P ∨ Q. In every valuation the implication holds, so it is a tautology.

• Third proposition: ((P ∨ ¬P) → Q) → ((P ∨ ¬P) → R). Since P ∨ ¬P is a tautology, each antecedent reduces to Q and R respectively, so the formula becomes Q → R, which is not a tautology in general.

• Fourth proposition: (Q → (P ∧ ¬P)) → (R → (P ∧ ¬P)). Because P ∧ ¬P is a contradiction, Q → (P ∧ ¬P) simplifies to ¬Q and R → (P ∧ ¬P) simplifies to ¬R, so the formula becomes ¬Q → ¬R (equivalently R → Q), which is not a tautology.

Therefore, only the first and second propositions are tautologies.