Some assistance with the following would be appreciated. This question seems to

Question

Some assistance with the following would be appreciated. This question seems to have been answered a few times albeit incorrectly. I am looking for all steps broken down with the reasons. Ex Commutative Law of Addition. I would appreciate any help and thumbs up will be given thanks.

Give reasons for the following steps in the proof sequence:

(x + y) + (y x’) = x + (y + (y x’))           

                        = x + ((y 1) + (y x’))     

                        = x + (y (1 + x’))           

                        = x + (y (x’ + 1))            

                        = x + (y + 1)                   

                      = x + y                             

Explanation / Answer

Ans.

(x+y)+(y.x')= x+y+ y.x'

= x+(y+(yx')) // we are making this syntax as per given question. Although there are many ways to reduce expression into x+y

= x+((y.1)+y.x') // Since A.1=A Identity law of Boolean algebra. So, y.1=y

=x+(y.(1+x')) // Distributive law A.(B+C)= A.B+ A.C . So, y.(1+x')=y.1+y.x'

= x+(y.(x'+1)) // Commutative Law A+B=B+A. So 1+x'=x'+1

= x+(y.1) // Annulment Law A+1=1. Therefore x'+1=1

= x+ y //Since A.1=A Identity law of Boolean algebra. So, y.1=y

Note: in question the line x + (y + 1) is incorrect because if we use laws of boolean algebra, then

x+(y+1)= x+1 // Annulment Law A+1=1. So, y+1=1

=1   // Annulment Law A+1=1. So, x+1=1 but correct answer is x+y.

So, x+(y+1) expression is incorrect in the given question.

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