# Consensus Theorem and Boolean algebra

I am trying to prove the following boolean equality. $$bc + abc + bcd + a’(d+c) = abc + a’c + a’d$$

I have simplified the left side to $bc + a'd + a'c$ by factoring out a $bc(1)$. However, I am not able to figure out where that extra $a$ comes from. And in any case, wouldn't my answer be the more simplified form?

• You are trying to prove an identity, not simplify an expression, so your answer is a proof, not "the more simplified form". Have you tried a truth table? – Paul Sinclair Apr 12 '16 at 17:58
• We were told to prove this using only boolean logic and not truth tables. And right, I am trying to prove it, but I'm not able to do so correctly. – Jonathan Apr 12 '16 at 18:05

Yes, your answer is the more simplified form. If Left and Right reduce to same expression, you have proved it.

So attempt to reduce the Right side of expression to Left.

Left expression: $$bc+abc+bcd+ \overline a(d+c)$$ $$bc (1+a+d) + \overline ad+\overline ac$$ $$bc + \overline ad + \overline ac$$

Right:

$$abc + \overline ad + \overline ac$$ $$abc + \overline ad + \overline ac (1+b)$$ $$abc + \overline ad + \overline ac + \overline abc$$ $$bc (a + \overline a) + \overline ad + \overline ac$$ $$bc + \overline ad + \overline ac$$

Edit...

And the question has nothing to do with consensus. See Laws and Theorems of Boolean Algebra.

$(X + Y) • (\overline X + Z) • (Y + Z) = (X + Y) • (\overline X + Z)$ [13a]

$X Y + \overline X Z + Y Z = X Y + \overline X Z$ [13b]

With consensus, third term (with Y and Z) is absorbed by first two.