if $a'b+cd'=0$, then prove that $ab+c'(a'+d')=ab+bd+b'd'+a'c'd$

Question

Let $a'b+cd'=0$, then prove that $$ab+c'(a'+d')=ab+bd+b'd'+a'c'd$$

I would like to know how to solve this expression, not able to make any headway. I have tried canonical form expansion and reduction but the terms in the if condition does not match and also not able to generate the terms in the given expression.

Hi! And welcome to MSE. I think that you are missing some relations. If you start considering $ab+c'(a'+d')$, you can not make appear $d$, since your unique relation $a'b+cd'=0$ does not contain any $d$, for example. Hence, I guess you have some missing relations. It could be between those $a,b,c,d$ and their respective $a',b',c',d'$. Let us know. Edit the post. — idriskameni, Jan 16 '19 at 10:42

score 0 · Answer 1 · answered Jan 16 '19 at 14:18

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A truth table shows that left-hand-side and righ-hand-side are in fact equivalent if the constraint is fulfilled (indicated by background color):

answered Jan 16 '19 at 14:18

Axel Kemper

4,943

if $a'b+cd'=0$, then prove that $ab+c'(a'+d')=ab+bd+b'd'+a'c'd$

1 Answers1