Let us consider the formula (pq)→[(p→q)→q]. Let us show that this formula is a tautology using the method by contradiction. Suppose that the formula is not tautology. Then there exists (p0,q0) such that ∣(p0q0)→[(p0→q0)→q0]∣=F. It follows from definition of implication that ∣p0q0∣=T,∣(p0→q0)→q0∣=F. It follows from definition of conjunction that ∣p0∣=∣q0∣=T.
On the other hand, we have ∣(p0→q0)→q0∣=(T→T)→T=T→T=T=F. This contradiction proves that the formula (pq)→[(p→q)→q] is a tautology.
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