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

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- Oct 12th 2010, 08:05 AMStaryNightBoolean Algebra Problem
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 - Oct 12th 2010, 08:14 AMAckbeet
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. - Oct 12th 2010, 09:25 AMStaryNight
Sorry, I meant to type

Prove a.b + a'.c = (a+c).(a'+b) - Oct 12th 2010, 09:27 AMAckbeet
That I'll buy. What techniques are available to you?

- Oct 12th 2010, 09:42 AMStaryNight
It basically says to do it using 'Boolean Algebra'. We've studied DeMorgan's theorem and K-maps.

- Oct 12th 2010, 10:07 AMAckbeet
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?

- Oct 12th 2010, 11:42 AMStaryNight
I've managed to prove it by firstly writing a truth table and finding the inverse fuction and then applying De Morgan and using absorbtion. However, from the question I was supposed to prove the identity entirely algebraically.

- Oct 12th 2010, 12:08 PMAckbeet
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.