As written, your equation is false. Set a = 0, b = 0, c = 1. Then ab + a'c = 0 + 1 = 1. However, ac(a'+b) = 0(a'+b) = 0.
Perhaps you have a typo in there somewhere?
[EDIT]: OP corrected ac(a'+b) to (a+c)(a'+b) for the RHS.
I've just started learning boolean algebra at university and am stuck by the following problem:
Prove a.b + a'.c = (a+c).(a'+b) // Corrected from previous typo
Could somebody please give me a push in the right direction.
Many Thanks
As written, your equation is false. Set a = 0, b = 0, c = 1. Then ab + a'c = 0 + 1 = 1. However, ac(a'+b) = 0(a'+b) = 0.
Perhaps you have a typo in there somewhere?
[EDIT]: OP corrected ac(a'+b) to (a+c)(a'+b) for the RHS.
Ok. I would probably go with the K-map method. In order to do that, you're going to need to expand out the RHS so that it's in disjunctive normal form. What do you get when you do that?
Here's one approach. Start with
a'b'c+a'bc+abc'+abc, and simplify.
However, it is also true that
a'b'c+a'bc+abc'+abc=(a'b'c'+a'bc'+ab'c'+ab'c)'.
Simplify this RHS.
It's the step
a'b'c+a'bc+abc'+abc=(a'b'c'+a'bc'+ab'c'+ab'c)'
that you must justify. It follows, ultimately, because of the law of the excluded middle. You've basically exhausted all 8 possible truth assignments of the variables.
I'm not sure how much sense that makes in my mind, but I know what I'm trying to say.