TAOCP 7.2.2.2 Exercise 218
The ternary operator $u,?,v:w$ has value $v$ when $u=1$ and value $w$ when $u=0$.
Section 7.2.2.2: Satisfiability
Exercise 218. [20] [20] Express the formula $(x \vee A) \wedge (\bar{x} \vee B)$ in terms of the ternary operator $u,?,v : w$.
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Solution
The ternary operator $u,?,v:w$ has value $v$ when $u=1$ and value $w$ when $u=0$. Consider the two possible values of $x$.
If $x=1$, then
$$ (x\vee A)\wedge(\bar{x}\vee B) =(1\vee A)\wedge(0\vee B) =1\wedge B =B. $$
If $x=0$, then
$$ (x\vee A)\wedge(\bar{x}\vee B) =(0\vee A)\wedge(1\vee B) =A\wedge1 =A. $$
Therefore the expression selects $B$ when $x=1$ and selects $A$ when $x=0$. This is exactly the meaning of the ternary expression $x,?,B:A$. Hence
$$ \boxed{(x\vee A)\wedge(\bar{x}\vee B)=x,?,B:A} $$