Can someone show me the steps of simplification for this Boolean expression?

(!A!B!CD) + (!AB!C!D) + (!AB!CD) + (!ABCD) + (A!B!CD) + (ABCD)

The expression is a disjunction of six minterms. Each of them can be merged with another by leaving out a literal which occurs in positive and inverted form in the minterms.


$$\bar{A}\bar{B}\bar{C}D \lor \bar{A}B\bar{C}D = \bar{A}\bar{C}D \land (\bar{B} \lor B) = \bar{A}\bar{C}D$$

The simplified expression:

$$\bar{B}\bar{C}D \lor \bar{A}B\bar{C} \lor BCD$$

This can be shown graphically in a Karnaugh-Veitch map which depicts minterms and their neighbor terms:

enter image description here

The six minterm cells can be grouped in three adjacent pairs to find the simplified expression. Note that several possible simplifications might exist. In this case, the only three-term cover for all six minterms is $1+9$, $4+5$ and $7+15$.


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