Simplify Boolean algebra
f(X,Y,Z)=(X’Y+XZ)(X+Y’)
Use the distributive properties:
(X’Y+XZ)(X+Y’)
=X(X’Y+XZ) + Y’(X’Y+XZ)
=(XX'Y + XXZ) + (X'YY' + XY'Z)
=(Y+XZ) + (X'+XY'Z) [XX'=YY'true, XX=X]
=Y + X' + XZ + XZY'
=Y + X' + XZ [XZ+XZY' = XZ (absorption)]
Now your turn to do some work:
check the above simplification using a truth table on the original expression, and the simplified version.
To simplify the Boolean expression f(X,Y,Z) = (X’Y + XZ)(X + Y’), we can use various Boolean algebra laws and rules to simplify each term step by step.
Step 1: Distributive Law
Apply the Distributive Law to expand the expression.
f(X,Y,Z) = X’Y * X + X’Y * Y’ + XZ * X + XZ * Y’
Simplifying this gives us:
f(X,Y,Z) = X’YX + X’YY’ + XZX + XZY’
Step 2: Complement Law
Apply the complement rule to simplify the complemented variables.
XY’ = X + Y’ (A variable OR its complement equals 1)
Using this rule, we can simplify our expression further:
f(X,Y,Z) = X’YX + 0 + XZX + XZY’
Step 3: Identity Law
Apply the Identity Law to eliminate the 0 term.
0 + X = X (A variable OR 0 equals the variable)
f(X,Y,Z) = X’YX + XZX + XZY’
Step 4: Distributive Law
Apply the Distributive Law once again to factor out common terms.
f(X,Y,Z) = X’YX + X(ZX + ZY’)
Step 5: Simplification
Combining the terms:
1. X’YX = 0 (A variable AND its complement equals 0)
2. ZX + ZY’ = Z(X + Y’) (Distributive Law)
Substituting these simplifications back into the expression, we get:
f(X,Y,Z) = 0 + Z(X + Y’)
Simplifying further:
f(X,Y,Z) = Z(X + Y’)
Thus, the simplified Boolean expression for f(X,Y,Z) = (X’Y + XZ)(X + Y’) is f(X,Y,Z) = Z(X + Y’).