Simplify the Boolean expression " (x+y).(x+z) " .
I think answer is " x+y.z " But i don't know how t get that.
You should use the De Morgan Law (A+B)=(A'.B'). It works this way:
(X+Y)=X'.Y' and (X+Z)=X'.Z'
By commutativity: (X+Y).(X+Z)=(X'.Y').(X'.Z')=X'.Y'.X'.Z'=X'.X'.Y'.Z'
By idempotence: X'.X'=X'
Then: X'.X'.Y'.Z'=X'.Y'.Z'=X'.(Y'.Z')
Calling: Y'.Z'=W
Then: X'.(Y'.Z')=X'.W'
By De Morgan: X'.W'=(X+W) (I)
Negating the affirmation: W'=Y'.Z' then W=(Y'.Z')'=Y'+Z'=Y.Z (II)
By (I) and (II): (X+Y).(X+Z)=X+(Y.Z)=X+Y.Z