Answer to Question #224910 in Discrete Mathematics for Swastika

Question

Is (p q) > I(p>q)> q] a tautology? Why or why not?

Answer 1

Let us consider the formula $(pq) \to [(p\to q)\to q].$ Let us show that this formula is a tautology using the method by contradiction. Suppose that the formula is not tautology. Then there exists $(p_0,q_0)$ such that $|(p_0q_0) \to [(p_0\to q_0)\to q_0]|=F.$ It follows from definition of implication that $|p_0q_0|=T, |(p_0\to q_0)\to q_0|=F.$ It follows from definition of conjunction that $|p_0|=|q_0|=T.$

On the other hand, we have $|(p_0\to q_0)\to q_0|=(T\to T)\to T=T\to T=T\ne F.$ This contradiction proves that the formula $(pq) \to [(p\to q)\to q]$ is a tautology.