Boolean Algebra Simplification

Question

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)

Answer 1

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.

Example:

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

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:

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 $7+15$.