---

Proof 2: Probability of biconditional

If \(p \iff q\), then \(\Pr(p) = \Pr(q)\).

See: biconditional

Proof. If \(p \iff q\), then \(p \lor \lnot q\) is a tautology per

\[\begin{array}{cc|c|l}
  p & q & p \iff q & p \lor \lnot q \\
  \hline
  0 & 0 & \boxed{1} & 0 \lor \lnot 0 = 0 \lor 1 = \boxed{1} \\
  0 & 1 & 0 & \cdots \\
  1 & 0 & 0 & \cdots \\
  1 & 1 & \boxed{1} & 1 \lor \lnot 1 = 1 \lor 0 = \boxed{1} \> ,
\end{array}
\]

and \(p\) is mutually exclusive with \(\lnot q\) per

\[\begin{array}{cc|c|l}
  p & q & p \iff q & \lnot q \\
  \hline
  \boxed{0} & 0 & 1 & \boxed{1} \\
  0 & 1 & 0 & \cdots \\
  1 & 0 & 0 & \cdots \\
  \boxed{1} & 1 & 1 & \boxed{0} \> , \\
\end{array}
\]

so, we have

\[\begin{align*}
  \Pr(p \lor \lnot q) & = 1 && \text{per axiom 2} \\
  \Pr(p) + \Pr(\lnot q) & = 1 && \text{per axiom 3} \\
  \Pr(p) + 1 - \Pr(q) & = 1 && \text{per proof 1} \\
  \Pr(p) + 1 & = \Pr(q) + 1 && \text{add \(\Pr(q)\)} \\
  \Pr(p) & = \Pr(q) && \text{subtract \(1\)}. \quad \blacksquare
\end{align*}
\]