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) = (xy'+w'z)(wx'+yz')

the Apostrophes represent a line over the letter that it is after

Explanation / Answer

Solution: F(w,x,y,z) = (xy'+w'z)(wx'+yz') Taking complement and applying demorgan law: =>F'(w,x,y,z) = [(xy'+w'z)(wx'+yz')]' =>F'(w,x,y,z) = (xy'+w'z)'+(wx'+yz')' =>F'(w,x,y,z) = ((xy')'.(w'z)')((wx')'+(yz')') Again using demorgan law: F'(w,x,y,z) = (x'+y+w+z')(w'+x+y'+z) Take complement again: =>F(w,x,y,z) = [(x'+y+w+z')(w'+x+y'+z)]' =(x'+y+w+z')'+(w'+x+y'+z)' =xy'w'z+wx'yz Ans

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