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I have to simplify A+C'+B'CD but I don't see how.

I had to deduce the expression starting from a logic diagram in which two AND gates were used for the B'CD part. Seeing the diagram all I can think of is to use one AND gate instead of two, but I don't see how the actual boolean expression can be simplified.

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  • $\begingroup$ Try to get the product of sum expression from K-map $\endgroup$ – Deep Joshi Apr 5 '17 at 9:59
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Using the property $X+YZ = (X+Y)(X+Z)$, you can silmplify $C' + B'CD$ as $C' + B'D$.

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