Can someone help me simplify this boolean expression? $$(a+b+c+d)(a'+b'+c'+d')$$
so if I use the distributive property, I'll get:
$$ab'+ac'+ad'+ba'+bc'+bd'+ca'+cb'+cd'+da'+db'+dc'$$
I'm stuck after this step...
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1$\begingroup$ I assume the result asks you to simplify the expression. $a+b+c+d=0$ exactly when $a=b=c=d=0$. $a'+b'+c'+d'=0$ exactly when $a=b=c=d=1$. In all other cases, both sums have value $1$. So when is their product $1$? $\endgroup$– Dilip SarwateCommented Feb 4, 2012 at 2:36
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1 Answer
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Dilip’s hint is good. Alternatively, if you know the de Morgan’s laws, you can observe that $$a'+b'+c'+d\,'=(abcd)'$$ and $$a+b+c+d=(a'b'c'd\,')'\;,$$ so that $$(a+b+c+d)(a'+b'+c'+d\,')=(a'b'c'd\,')'(abcd)'\;.\tag{1}$$ Now use de Morgan again to get rid of the outer negations on the righthand side of $(1)$.
answered Feb 4, 2012 at 9:38