Boolean algebra

[xoxpe]

I dont know where i should put this question...please forgive me if i put this in wrong place...

 

i have no idea how to prove this boolean algebra equation are equivalent:

1. x'y' + xy = (xy' +x'y)'
   
2. x'z + xy = x'y'z + yz + xy
   

 

thank u very much for your help...

[lqg]

x'y' + xy = (xy' +x'y)'

x'z + xy = x'y'z + yz + xy

 

if (xy'+x'y)=1

then either xy'=1 x'y=0 or xy'=1 x'y=0 or xy'=xy'=1

1) xy'=1 x'y=0 then x=y'=1,

obviously x'y'=0 and xy=0 cause y=0 and x'=0. and (xy'+x'y)'=0 so we have an equality.

this way you prove for the other two cases and for the case where (xy'+x'y)=0 which is another three cases.

[uncool]

1)I assume ' means opposite.

(xy' + x'y)' = (xy')'(x'y)' = (x' + y)(x + y') = xx' + x'y' + xy + yy' = 0 + xy + x'y' + 0 = xy + x'y'

Laws used here: Demorgan's, Demorgan's, Distributive, Contradiction, Exclusion (I htink the last one is right, not sure)

 

 

2)x'z + xy = x'y'z + yz + xy

x'z + xy = x'yz + x'y'z + xy = x'y'z + y(x + x'z) = x'y'z + y(x + z) = x'y'z + yz + xy

Laws used here: 1) Can't remember what it's called, distributive, can't remember, distributive.

=Uncool=