Electrical – How to simply this 2 to 1 mux boolean expression

boolean-algebramux

I am going through this tutorial for a 2 to 1 mux.
They create this circuit:
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They then derive this boolean algebra expression and simplification:
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I'm confused how they made the simplification though. I see that the expression can be simplified to \$I_0*I_1 * (\overline{A} + A)\$ by simplifying the double negatives, but I don't see how they got any farther than that

Best Answer

I use a slightly different notation/symbols to denote it easier:

  • a = not A
  • A = A
  • b = not I1
  • B = I1
  • c = not I0
  • C = I0
Q = aBc + aBC + AbC + ABC   
  = aB(c + C) + AC(b + B)     // c + C = true, aB(true) = aB
                              // b + B = true, AC(true) = AC
  = aB        + AC

Written back in original symbols:

Q = aB + AC = not(A) * I1 + A * I0