Converse, Inverse, Biconditional

Converse and Inverse

$\begin{matrix} p & q & p \to q & q \to p& \neg p & \neg q & \neg p \to \neg q\\ T & T & T & T & F & F & T\\ T & F & F & T & F & T & T\\ F & T & T & F & T & F & F\\ F & F & T & T & T & T & T \end{matrix}\\ so,\ converse\ \equiv\ inverse$

Biconditional Statement

$p \iff q$

if and only if (iff)

$\begin{matrix} p & q & p \iff q \\ T & T & T \\ T & F & F \\ F & T & F \\ F & F & T \end{matrix}$

$ex1\\ p \iff q \equiv (p \to q) \land (q \to p)\\ \begin{matrix} p & q & p \to q & q \to p & (p \to q) \land (q \to p) & p \iff q \\ T & T & T & T & T & T \\ T & F & F & T & F & F \\ F & T & T & F & F & F \\ F & F & T & T & T & T \end{matrix}$

$ex2\\ (p \to (q \to r)) \equiv ((p \land q) \to r)\\ proof\\ \begin{matrix} p & q & r & q \to r & (p \to (q \to r) & p \land q & (p \land q) \to r\\ T & T & T & T & T & T & T \\ T & T & F & F & F & T & F \\ T & F & T & T & T & F & T \\ T & F & F & T & T & F & T \\ F & T & T & T & T & F & T \\ F & T & F & F & T & F & T \\ F & F & T & T & T & F & T \\ F & F & F & T & T & F & T \\ \end{matrix}$