Question

proove (x+y)'z+xy'=y'(x+z) of Boolean alzebra​

Answer 1

Answer:

Theorem 3(a) and (b)

THEOREM 3(a) Law of Absorption : yx+x = x.

Proof :

yx+x = yx+x1 by identity (Ax. 2b)

= x(y+1) by distributivity (Ax. 4a)

= x1 by Theorem 2(a)

= x by identity (Ax. 2b)

THEOREM 3(b): x(x+y) = x by duality.

Theorem 6(a) and (b)

THEOREM 6(a) De Morgan's Laws: (x+y)' = x'y'.

Proof: We will prove that x'y' is a complement of x+y by proving that x'y' satisfies Axiom 5, which states that x+x' = 1 and that xx' = 0. Thus we have to prove that (a) (x+y)+(x'y')=1 and (b) (x+y)(x'y') = 0.

Part (a):

(x+y)+(x'y') = x+(y+(x'y')) by assiociativity (Th. 5a)

= x+((x'y')+y') by commutativity (Ax. 3a)

= (x+(x'y'))+y by assiociativity (Th. 5a)

= (x+x')(x+y')+y by distributivity (Ax. 4b)

= 1(x+y')+y by complement (Ax. 5a)

= (x+y')+y by identity (Ax. 2b)

= x+(y'+y) by assiociativity (Th. 5a)

= x+1 by complement(Ax. 5a)

= 1 by Theorem 2(b)

Part (b):

(x+y)(x'y') = x(x'y')+y(x'y') by distributivity (Ax. 4a)

= x(x'y')+y(y'x') by commutativity (Ax. 3b)

= (xx')y'+(yy')x' by associativity (Th. 3b)

= 0y'+0x' by complement(Ax. 5b)

= 0+0 by identity(Ax. 2b)

= 0