s x_{1} x_{0}  z s  z x_{1}x_{0 }0 0 0  0 0  x_{0} s \00 01 11 10 0 0 1  1 1  x_{1} 0 0 1 1 0 0 1 0  0 1 0 0 1 1 0 1 1  1 1 0 0  0 z = x_{1}s + x_{0}s' 1 0 1  0 1 1 0  1 1 1 1  1The gate network to implement a two input mux and symbol is:
NOT
z = 0x_{1}+ 1x_{1}' = x_{1}' 

AND
z = x_{1}x_{0}+ 0x_{0}' = x_{1}x_{0} 

OR
z = 1x_{0}+ x_{1}x_{0}' = x_{1}+x_{0} 
From the text example,
z = x_{3}(x_{1}+x_{2}x_{0}) = x_{3}x_{1}+x_{3}x_{2}x_{0}
f(x_{3},x_{2},x_{1},0) = x_{3}x_{1}x_{0}^{'}+x_{3}x_{2}0
= x_{3}x_{1}1 + 0 = x_{3}x_{1}x_{0}^{'}
f(x_{3},x_{2},x_{1},1) = x_{3}x_{1}x_{0}+x_{3}x_{2}x_{0}
= x_{3}x_{1}1+x_{3}x_{2}1_{ }=
x_{3}x_{1}x_{0}+ x_{3}x_{2}x_{0}
so
z = x_{3}x_{1}x_{0}' + _{ }x_{3}x_{1}x_{0}+
x_{3}x_{2}x_{0} = x_{3}x_{1}x_{0}'
+ _{ }x_{3}(x_{1}+x_{2})x_{0}
This was implemented by:
The algebraic expression for a fourinput multiplexer is: z = x_{3}s_{1}s_{0}+ x_{2}s_{1}s_{0}'+x_{1}s_{1}'s_{0}+x_{0}s_{1}'s_{0}' which would have 6 inputs and 64 rows for a truth table solution. The functional behavior for the 4input multiplexer and modular implementation would be:
s1 s0  Output
0 0  x0 0 1  x1 1 0  x2 1 1  x3 