Asked by Ardi
Simplify Boolean algebra
f(X,Y,Z)=(X’Y+XZ)(X+Y’)
f(X,Y,Z)=(X’Y+XZ)(X+Y’)
Answers
Answered by
MathMate
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.
(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.
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