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I want to simplify the below equation but I can't get the same result with an online calculator (the result for the calculator is the correct - I check it with Logisim).

$Y = A'B'CD + A'B'C'D + ABC'D' + A'BC'D + AB'C'D'$

$Y = A'B'(C'D + CD) + A'BC'D + C'D'(AB' + AB)$

--> From Redundancy Laws: $AB + AB' = A$

$Y = A'B'D + A'BC'D + C'D'A$

--> I can simplify until here

The complete simplification is the below, but I can't find a boolean's law to use to remove the $B$.

Does anyone knows which law I have to use in order to remove the variable?

$Y = A'B'D + A'C'D + C'D'A$

Bram28
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1 Answers1

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Here is something you can do:

$A'B'D + A'BC'D + C'D'A =$

$A'D(B' + BC') + C'D'A =$

$A'D(B' + B)(B'+C') + C'D'A =$

$A'D1(B'+C') + C'D'A =$

$A'D(B'+C') + C'D'A =$

$A'B'D + A'C'D + C'D'A$

Bram28
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