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language-agnosticbooleanboolean-logicboolean-expressionboolean-operations

How was this boolean expression further simplified?


(ab+cd)(a'b'+c'd') = 1+ abc'd' + a'b'cd +1

so I'm stuck at

abc'd'+a'b'cd 

but the final answer is

(a+b)(c+d)+(a'+b')(c'+d')

What am I missing?


Solution

  • It seems to me that those two expressions are complementary, i.e. the only two cases where (a+b)(c+d)+(a'+b')(c'+d') are false are abc'd' and a'b'cd.

    Edit: Somewhere along the line I think you've lost a ' and you're actually looking for one of these:

    ((ab+cd)(a'b'+c'd'))'
    (ab+cd)'+(a'b'+c'd')'
    ((ab)'(cd)')+((a'b')'(c'd')')
    (a'+b')(c'+d')+(a+b)(c+d)
    (a+b)(c+d)+(a'+b')(c'+d')
    
    (ab+cd)(a'b'+c'd')
    (a'b'+c'd')(ab+cd)
    ((a+b)'+(c+d)')((a'+b')'+(c'+d')')
    ((a+b)(c+d))'((a'+b')(c'+d'))'
    ((a+b)(c+d)+(a'+b')(c'+d'))'