# How to simply this 2 to 1 mux boolean expression

I am going through this tutorial for a 2 to 1 mux. They create this circuit:

They then derive this boolean algebra expression and simplification:

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

• Do you know about Karnaugh maps? If not, google that. – The Photon Aug 18 '17 at 1:41
• Another way to arrive at that final formula is to look at the circuit. The final NAND has all its inputs and outputs inverted, so you can replace it (and all the inversions) by an OR. That's the + in your final formula. The two items that are added are A*I1 and not(A)*I2, so there you are. – Wouter van Ooijen Aug 18 '17 at 10:25
• @WoutervanOoijen I don't understand what you're saying. What do you mean you can just replace the final NAND and it's inputs with an OR? Wouldn't that give not NOT(I0&A) OR NOT (NOT A& I1)? – Tyler H Aug 18 '17 at 18:54
• A or B == not( not( A ) AND not( b ) ) – Wouter van Ooijen Aug 18 '17 at 19:54

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