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I want to minimize boolean equation, but I have no idea how to do. Here is the equation

Y = ABC~D + A~(BCD) + ~(A+B+C+D).

Can it be minimized?

Green.H
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Aiden
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  • The usual way is to write a truth-table. For $4$ input variables, it has $2^4 = 16$ rows called minterms. In a second step, fill the $16$ truth values in a Karnaugh Veitch map to find an expression with a minimal number of implicants. – Axel Kemper May 03 '18 at 08:33

1 Answers1

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It seems to me you're using $\sim$ as the Boolean complement with prefix notation.
So with a more standard notation your expression is $$Y=ABCD'+A(BCD)'+(A+B+C+D)'.$$ This expression can be simplified in different ways, and it is not clear to me which one you'd like.
Here goes its DNF form: \begin{align} Y &= ABCD'+A(B'+C'+D')+A'B'C'D' \tag{de Morgan laws}\\ &= A(BCD'+B'+C'+D')+A'B'C'D' \tag{distributuvity}\\ &= AB'+AC'+AD'+A'B'C'D' \tag{absorption & distrib} \end{align} where absorption comes from $BCD'+D'=D'$.
Now \begin{align} AB' + A'B'C'D' &= B'(A+A'C'D') \tag{distributivity}\\ &= B'((A+A')(A+C')(A+D')) \tag{distributivity}\\ &= B'(A+C')(A+D') \tag{since $A+A'=1$}\\ &= B'(A+C'D') \tag{distributivity}\\ &= AB' + B'C'D', \tag{distributivity} \end{align} whence $$Y=AB'+AC'+AD'+B'C'D'.$$

From here it also follows that $$Y=A(BCD)'+(B+C+D)',$$ which is obtained from the original one by erasing the first summand and also the first summand of the complement of the last one. I suspect this is really what you want.

amrsa
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