√完了しました！ p^q q^p 142208-P(q(x))=q(p(x))

Here are your steps Subtract q from both sides if positive, or add q to both sides if negative This gives you px q q = r q, which simplifies to px = r q Divide by p on both

P(q(x))=q(p(x))-If P and Q are two premises, we can use Conjunction rule to derive $ P \land Q $ $$\begin{matrix} P \\ Q \\ \hline \therefore P \land Q \end{matrix}$$ Example Let P − "He studies very hard" Let QWritten p q or p q, iff the compound proposition p q is a tautology Compound propositions p and q are logically equivalent to each other iff p and q contain the same truth values as each other in all