Simplify the following functional expressions using boolean algebra and its iden

Question

Simplify the following functional expressions using boolean algebra and its identities. List the identity used at each step

F(w,x,y,z) = (x+y)'(x'+y')'

the Apostrophes represent a line over the letter that it is after. The last set of parenthesis has a line over it as well as the x and y in them.

Explanation / Answer

Solution: F(w,x,y,z) = (x+y)'(x'+y')' =>F'(w,x,y,z) = [(x+y)'(x'+y')']' Using de morgan law: F'(w,x,y,z) = (x+y)+(x'+y') =>F'(w,x,y,z) = (x+y)+(x'+y') =>F'(w,x,y,z) = x+y+x'+y' using property x+x'=1 =>F'(w,x,y,z) = 1+1 =>F'(w,x,y,z) = 1 =>F'(w,x,y,z) = 0 Ans Please rate

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