Contrapositive equivalence

We prove $(P \to Q) \equiv (\neg Q \to \neg P)$ by comparing truth tables for $P\to Q$ and $\neg Q\to\neg P$ across all four assignments.

Case $P=1, Q=1$: $P\to Q=1$ (a true hypothesis with a true conclusion). Also $\neg Q=0$, so $\neg Q\to\neg P$ has a false hypothesis, which makes the whole conditional $1$ regardless of $\neg P$. Both sides are $1$.

Case $P=1, Q=0$: $P\to Q=0$ (true hypothesis, false conclusion is the only case that makes a conditional false). Also $\neg Q=1$ and $\neg P=0$, so $\neg Q\to\neg P$ has a true hypothesis and false conclusion, giving $0$. Both sides are $0$.

Case $P=0, Q=1$: $P\to Q=1$ (false hypothesis makes any conditional true). Also $\neg Q=0$, so $\neg Q\to\neg P=1$ automatically. Both sides are $1$.

Case $P=0, Q=0$: $P\to Q=1$ (false hypothesis). Also $\neg Q=1$, $\neg P=1$, so $\neg Q\to\neg P=1$ (true hypothesis, true conclusion). Both sides are $1$.

All four rows agree, so $P\to Q$ and $\neg Q\to\neg P$ have identical truth tables and are logically equivalent. Note this differs from the converse $Q\to P$ and the inverse $\neg P\to\neg Q$, which are equivalent to each other but generally NOT equivalent to the original conditional — a common source of invalid arguments.