Simplifying Gates

Question

I have a practice question that I'm working on but I can't seem to figure out how to simplify ABC'+ C. The answer is supposed to be AB + C. I'm not sure how to get that answer. I tried to expand C into ABC' and simplified from there, but I ended up with AB + AC + BC + C. I can easily see how to get AB + C by drawing a schematic and just using NAND gates but I don't know how to simplify it "algebraically". Thanks!

Answer 1

I figured it out. I had to take it a bit further.

ABC' + C (distribute C)

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

=(A+C)(B+C)(1) (expand)

=AB+AC+BC+CC

=[C(A+B+1)] + AB

=C(1) + AB

=AB + C

Answer 2

Probably the simplest derivation is by distributing the $C$ over $AB$ and $C'$:

$$ABC'+C = (AB+C)(C'+C) =(AB+C)1=AB+C$$

Then again, this pattern is so common that some texts will have this as a basic equivalence:

Reduction

$P+P'Q=P+Q$

So with Reduction it would just be $1$ step: Reduction!