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Simplify A'BC'+ABD+A'C+A'CD'+AC'D+AB'C'
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Well first the 3rd and 4th term is in the form of X+XY = X(1+Y) = X so first you can simplify easily A'BC'+ABD+A'C+AC'D+AB'C'
A'BC'+A'C+ABD+AC'D+AB'C' A'(BC'+C)+A(BD+C'D+B'C')
@amistre64
A(BD+B'C'+C'D) Within the brackets is consensus if X=B, Y=C' , Z=D XZ+X'Y + YZ = XZ+X'Y
im not too sure what a simplified version of it would look like
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so now we got A'(BC'+C)+A(BD+B'C')
I dont see anything else @amistre64
A'BC'+A'C + ABD+AB'C'
@amistre64 My professor has the answer as A'C + A'D' + ABD
so somehow he cancels the other two where you only have a D' remaining
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