Prove $ab + ab\overline{c} + bcd = b(a+c)(a+d)$

Question

Do I need to use absorbtion law to prove them?

1. $ab + ab\overline{c} + bcd = b(a+c)(a+d)$
2. $ab + cd = (a+c)(a+d)(b+c)(b+d)$.

For 1), I simplified $ab+ ab\overline{c} + bcd$ into $b(a\overline{c} + cd)$, then I get stuck.

For 2), I found that if I expand $(a+c)(a+d)(b+c)(b+d)$, I will not be able to obtain $ab+cd$, what should I do?

Answer 1

Hint: Express both sides of the equation in form of minterms (terms containing all variables). This can be done by multiplying by $(x+\bar{x})$ where $x$ is the missing variable in a term.

Answer 2

given $$abc +abc'+bcd$$ $$ab(1+c') +bcd$$ since $1+c'=1$$$ab+bcd$$ $$b(a+cd)$$ using the distributive law $a+cd=(a+c)(a+d)$ gives you $$b(a+c)(a+d)$$