Prove 2 Boolean expressions are equal by multiplying out and simplifying

Question

I was asked to prove that the following is TRUE by "multiplying it out" and simplifying:

$(A+BC)(A+DE)=A+BCDE$

I am already familiar with the theorem

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

which would be enough proof to prove this true in standard environments, but I was specifically told to "multiply it out". So I tried and this is how far I got:

$(A+BC)(A+DE)$

$=AA+BCDE+ABC+ADE$

$=A+BCDE+ABC+ADE$

$=A(BC+DE)+BCDE$

How does that look so far? Am I headed in the right direction, or do I need to take a different route for simplification?

This is the distributivity of addition over multiplication. In general, we have $(a+b)c = ac+bc$ as a law over the real numbers, right? This means that multiplication distributes over addition. In boolean algebra, we have the other rule as well, which is definitely not true over the real numbers. This rule is $(a+b)(a+c) = a + bc$, which looks weird, but is true because $a+bc = a(1+b+c) + bc = (a+b)(a+c)$. — Sarvesh Ravichandran Iyer, Sep 20 '17 at 02:35

score 2 · Accepted Answer · answered Sep 20 '17 at 02:16

2

You need to take a different route at your second-to-last line: $$A+BCDE+ABC+ADE$$ $$=A(BC+DE+1)+BCDE$$ $$=A+BCDE$$

answered Sep 20 '17 at 02:16

Parcly Taxel

103,344

Thank you. How does A(BC+DE+1) simplify to A? I think there's a theorem I'm unfamiliar with here. – wad11656 Sep 20 '17 at 02:21
@wad11656 In $BC+DE+1$ the 1 makes the whole expression true regardless of what $B,C,D,E$ are. So $A(BC+DE)=A1$, and the latter reduces to $A$ by identity of AND. – Parcly Taxel Sep 20 '17 at 02:26
Ah, perfect. Thanks so much – wad11656 Sep 20 '17 at 02:27

Prove 2 Boolean expressions are equal by multiplying out and simplifying

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