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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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    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 $1$?2012-02-04

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