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...
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...
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 $(1)$.