Prove, not using truth table, that $a b+a^{\prime} b^{\prime}+b c=a b+a^{\prime} b^{\prime}+a^{\prime} c$

Question

Question: Let $B\left(+, \cdot,^{\prime}\right)$ be a Boolean algebra and $a, b, c \in B$. Prove, not using truth table, that $a b+a^{\prime} b^{\prime}+b c=a b+a^{\prime} b^{\prime}+a^{\prime} c$.

Sol: We have $a b+a^{\prime} b^{\prime}+b c=ab+a'b'+bc(a+a')$.
I stuck here.

Answer 1

You're on the right track. Expand to get $$ab + a'b' + abc + a'bc$$ Then group the terms again $$ab(1+c) + a'(b' + bc)\\ = ab + a'b' + a'c$$ Last step uses the property $b' + bc = b'c + b' + bc = b' + c$, which is called absorption law AFAIK.

Answer 2

$ab+a'b'$ is true iff $a=b$, and when they are different, $bc=a'c$.

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More formally,

• $b=a\to aa+a'a'+ac=\text{true}=aa+a'a'+a'c$,

• $b=a'\to aa'+a'a+a'c=aa'+a'a+a'c$.

Answer 3

You're almost done.

$ab+a'b'+bc(a+a')= (ab+abc)+a'b'+a'bc = ab(1+c)+a'(\color{blue}{b'+bc}) = ab +a'(\color{blue}{b'+c})$

$\Rightarrow \color{red}{ab+a'b'+bc= ab+a'b'+a'c}$

For the blue part, $b'+bc = (b'+b)(b'+c) = 1(b'+c)=b'+c $