The following propositional statement is (P → (Q v R)) → ((P ^ Q) → R)

Answer: the formula is satisfiable but not valid.

Counterexample showing the formula is not valid:

• Take P = true, Q = true, R = false.

  Then P → (Q v R) = true → (true v false) = true → true = true,

  (P ∧ Q) → R = (true ∧ true) → false = true → false = false,

  so the whole implication (P → (Q v R)) → ((P ∧ Q) → R) = true → false = false.

This single assignment makes the formula false, so the formula is not valid.

Example showing the formula is satisfiable:

• Take P = false, Q = false, R = false.

  Then P → (Q v R) = false → false = true, and (P ∧ Q) → R = (false ∧ false) → false = false → false = true, so the whole formula is true.

Since the formula is true for at least one assignment but false for another, it is satisfiable but not valid.