Boolean Simplification: AB + BC + ~AC

Question

I have this logic equation: $$AB + BC + \sim AC$$

I know that the simplified equation is $AB + \sim AC$, but I don't know the steps to get there. Can anyone help me?

Answer 1

Quick & easy ... Do a truth table \begin{array}{l|l} A & B & C & AB & BC & (AC)'& A'+B+C' \\ \hline 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 1 \\ \end{array}

Alternatively, the algebriac proof uses De Morgans law $(AC)'=A'+C'$ and $A'+AB=B$ and $BC+C'=B$ and $B+B=B$.

Answer 2

I am wondering how to read your use of ~AC. Is this supposed to be ~(AC) or (~A)C?

Let's first see what happens if we interpret it as ~(AC) ... which I will write as (AC)', as that is more common in this kind of notation:

$$AB + BC + (AC)'=$$

$$AB + BC + A' + C'=$$

$$AB + A' + BC + C'=$$

$$(A + A')(B + A') + (B+C')(C + C')=$$

$$1(B + A') + (B+C')1=$$

$$B + A' + B+C'=$$

$$B + A' +C'=$$

$$B + (AC)'$$

Hmm, clearly this is not your desired AB + ~BC

OK, let's treat your ~AC as (~A)C ... which again I will write as A'C

Then we can do:

$$AB + BC + A'C=$$

$$AB + 1BC + A'C=$$

$$AB + (A + A')BC + A'C=$$

$$AB + ABC + A'BC + A'C=$$

$$AB(1+C) + A'C(B + 1)=$$

$$AB1 + A'C1=$$

$$AB + A'C$$

Aha! So that is your desired AB + ~AC ... if we interpret ~AC as (~A)C.