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Proof 1: Probability of negation

\[\begin{equation*}
  \Pr(\lnot p) = 1 - \Pr(p).
\end{equation*}
\]

See: negation

Proof. \(p \lor \lnot p\) is a tautology per

\[\begin{equation*}
  p \lor \lnot p =
  \begin{cases}
    0 \lor \lnot 0 = 0 \lor 1 = \boxed{1} & p = 0 \\
    1 \lor \lnot 1 = 1 \lor 0 = \boxed{1} & p = 1, \\
  \end{cases}
\end{equation*}
\]

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

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

so, we have

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