Basic boolean prove

Question

I need to prove that given $$ f_1 = c + a'd' + bd' \quad\text{and}\quad f_2 = a'b'd' + a'bd' + ab'c + abd' $$ that $f_1 = f_2$. How do I manipulate $f_2$ to be exactly like $f_1$? I have tried a lot with no success. Thanks.

Answer 1

Assuming that $+$, $\cdot$ and $'$ stand for the logical OR, AND and NOT operations respectively, then $f_1 \neq f_2$. Try for example $(a,b,c,d) = (0,0,1,1)$.

Tip: When you fail to prove something, try to disprove it. You will either succeed or hopefully get an idea why you can't (so then you can go on and prove it).

Answer 2

Consider $$ f_1d = (c + a'd' + bd')d = cd+a'd'd+bd'd=cd $$ However $$ f_2d = (a'b'd' + a'bd' + ab'c + abd')d = ab'cd $$ For $a=0$, $c=d=1$ we have $f_1d=1$, whereas $f_2d=0$.

So, of course, $f_1\ne f_2$, unless we're in the trivial boolean algebra where $0=1$.