Biconditional Equivalent to Biconditional of Negations/Formulation 1/Reverse Implication

Theorem

 * $\neg p \iff \neg q \vdash p \iff q$

Proof

 * align="right" | 2 ||
 * align="right" | 1
 * $\left ({\neg p \implies \neg q}\right) \land \left ({\neg q \implies \neg p}\right)$
 * By definition
 * 1
 * align="right" | 3 ||
 * align="right" | 1
 * $\left ({\neg \neg q \implies \neg \neg p}\right) \land \left ({\neg \neg p \implies \neg \neg q}\right)$
 * TP
 * 2
 * TP
 * 2


 * align="right" | 5 ||
 * align="right" | 1
 * $\left ({p \implies q}\right) \land \left ({q \implies p}\right)$
 * Comm
 * 4
 * align="right" | 6 ||
 * align="right" | 1
 * $p \iff q$
 * By definition
 * 5
 * By definition
 * 5