Use properties of Boolean algebra to simplify the following Boolean expression (showing all the steps):
(x + y′)(x′ + y′)′
Use properties of Boolean algebra let us simplify the following Boolean expression:
(x+y′)(x′+y′)′=(x+y′)(x′′⋅y′′)=(x+y′)(x⋅y)=x⋅x⋅y+y′⋅x⋅y=x⋅y+y′⋅y⋅x=x⋅y+0⋅x=x⋅y+0=x⋅y(x + y')(x' + y')'= (x + y')(x'' \cdot y'')= (x + y')(x\cdot y)\\= x\cdot x\cdot y + y'\cdot x\cdot y= x\cdot y + y'\cdot y\cdot x= x\cdot y +0\cdot x\\= x\cdot y +0=x\cdot y(x+y′)(x′+y′)′=(x+y′)(x′′⋅y′′)=(x+y′)(x⋅y)=x⋅x⋅y+y′⋅x⋅y=x⋅y+y′⋅y⋅x=x⋅y+0⋅x=x⋅y+0=x⋅y
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