Simplify the Boolean expression M = W'XY' + WXY' + XY'Z, where an apostrophe denotes complement.

Step-by-step solution

1. 1

   Look for a product that every term shares. Term by term the literals are $\{\overline{W},X,\overline{Y}\}$, $\{W,X,\overline{Y}\}$ and $\{X,\overline{Y},Z\}$. The pair $X\overline{Y}$ appears in all three, so distributivity lets us pull it out rather than build a Karnaugh map: $M=X\overline{Y}\left(\overline{W}+W+Z\right).$

2. 2

   Apply the complement law inside the bracket. For any variable, $\overline{W}+W=1$, because at least one of a variable and its complement is always true. Substituting, $M=X\overline{Y}\left(1+Z\right).$

3. 3

   Apply the annulment law. In Boolean algebra OR-ing anything with $1$ gives $1$: $1+Z=1.$ This is where Boolean algebra parts company with ordinary arithmetic - there is no carry, and $1$ absorbs everything.

4. 4

   Finish the simplification. $M=X\overline{Y}\cdot 1=X\overline{Y}.$ Three three-literal minterms collapse to a single two-literal product term, so the circuit needs just one inverter and one AND gate instead of three AND gates and an OR gate.

5. 5

   Verify with a truth table. $M$ should be $1$ exactly when $X=1$ and $Y=0$, for every $W$ and $Z$. Checking the original: with $X=1,Y=0$ the three terms are $\overline{W}$, $W$ and $Z$, and $\overline{W}+W+Z=1$ always. With $X=0$ every term contains the factor $X=0$, so $M=0$; with $Y=1$ every term contains $\overline{Y}=0$, so again $M=0$. All 16 rows of $(W,X,Y,Z)$ agree with $X\overline{Y}$.