Help: how to proof boolean algebraic identity

Question

I am stuck on this question.

Proof the following:$$(abd+a’b+b’d+c’)(c+ab+bd) = b(a+c)(a’+c’)+d(b+c)$$

So far I could only figure out the following.

LHS: $(abd+a’b+b’d+c’)(c+ab+bd)$

1. Expand everything: $$(abcd+aabbd+abbdd)+(a'bc+aa'bb+a'bbd)+(b'cd+abb'd+bb'dd)+(cc'+abc'+bc'd)$$

2. Simplify idempotent terms: $$(abcd+abd+abd)+(a'bc+0 \cdot b+a'bd)+(b'cd+a \cdot 0 \cdot d+0 \cdot d)+0+abc'+bc'd$$ $$=abcd+abc'+abd+a'bc+a'bd+bc'd+b'cd$$

3. Factor out $b$ ($bc'd$ is left out cause it doesn't help): $$b(acd+ac'+ad+a'c+a'd)+bc'd+b'cd$$

4. Absorption ($acd+ad=ad$ and $ad+a'd=d$): $$b(ac'+a'c+d)+bc'd+b'cd$$

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RHS: $b(a+c)(a’+c’)+d(b+c)$

1. Expand everything: $$b(aa'+ac'+a'c+cc')+bd+cd$$ $$=aa'b+abc'+a'bc+bcc'+bd+cd$$

2. Simplify idempotent terms: $$0 \cdot b+abc'+a'bc+b \cdot 0+bd+cd$$ $$=abc'+a'bc+bd+cd$$

Not sure where I can go from here. Any help is greatly appreciated!

Answer 1

Almost there!

Let's expand what you ended up with the LHS:

$b(ac'+a'c+d)+bc'd+b'cd=$

$abc'+a'bc+bd+bc'd+b'cd$

Absorption ($bd+bc'd=bd$)

$abc'+a'bc+bd+b'cd$

Factor

$abc'+a'bc+(b+b'c)d$

Distribution

$abc'+a'bc+(b+b')(b+c)d$

$abc'+a'bc+1(b+c)d$

$abc'+a'bc+(b+c)d$

$abc'+a'bc+bd+cd$