---

Derive

1. Define probabilities of events \(A\) and \(B\) as

   \[\begin{equation*}
     \begin{array}{r|cc}
       & B & \lnot B \\ \hline
       A & s & t \\ \lnot A & u & v.
     \end{array}
   \end{equation*}
   \]

2. Substitute \(s, t, u, v\) into the law

   \[\begin{equation*}
     \begin{array}{ccccccccc}
       P(A) & = & P(A | B) & \cdot & P(B) & + & P(A | \lnot B) & \cdot & P(\lnot B)
       \\[3ex]
       \begin{array}{cc} \boxed{s} & \boxed{t} \\ u & v \end{array}
       & = & \begin{array}{cc} \boxed{s} & \bcancel{t} \\ u & \bcancel{v} \end{array}
       & \cdot & \begin{array}{cc} \boxed{s} & t \\ \boxed{u} & v \end{array}
       & + & \begin{array}{cc} \bcancel{s} & \boxed{t} \\ \bcancel{u} & v \end{array}
       & \cdot & \begin{array}{cc} s & \boxed{t} \\ u & \boxed{v} \end{array}
       \\[3ex]
       \dfrac{s + t}{s + t + u + v}
       & = & \dfrac{s}{s + u} & \cdot & \dfrac{s + u}{s + t + u + v}
       & + & \dfrac{t}{t + v} & \cdot & \dfrac{t + v}{s + t + u + v}.
     \end{array}
   \end{equation*}
   \]

3. Verify the equality

   \[\begin{align*}
       \frac{s + t}{s + t + u + v}
       & \stackrel{?}{=}
       \frac{s}{s + u}
       \cdot
       \frac{s + u}{s + t + u + v}
       +
       \frac{t}{t + v}
       \cdot
       \frac{t + v}{s + t + u + v}
       \\[2ex]
       \frac{s + t}{s + t + u + v}
       & =
       \frac{s \cancel{(s + u)}}{\cancel{(s + u)} (s + t + u + v)}
       +
       \frac{t \cancel{(t + v)}}{\cancel{(t + v)} (s + t + u + v)}
       \\[2ex]
       \frac{s + t}{s + t + u + v}
       & =
       \frac{s}{s + t + u + v}
       +
       \frac{t}{s + t + u + v}
       \\[2ex]
       \frac{s + t}{s + t + u + v}
       & =
       \frac{s + t}{s + t + u + v}.
       \\[2ex]
       1
       & \stackrel{\checkmark}{=}
       1.
   \end{align*}
   \]