CODEWITHSUNDEEP
boolean-logicboolean-expression
Remixt
Remixt

Reputation: 597

Is this simplification correct? (Demorgans Theorem)

{a(b+c)+a’b}’ 

using demorgans theorem I got a'+ b'c'a + b' then I factored b' out of b'c'a + b' to get b'(1+c'a) which just turns into b'. plugging it back into the equation I'm left with a'+b'. Is that correct or do I have this all wrong?

 

Upvotes: 0

Views: 44

Answers (1)

itzmebibin
itzmebibin

Reputation: 9439

{a(b+c)+(a'b)}' = (a (b+c))' . (a'b)'    
                = (a' + (b+c)') . (a+b')
                = (a' + (b'.c')) . (a+b')
                = (a.a') + (a'b') + (ab'c') + (b'c')
                = 0 + a'b' + b'c'(a+1)
                = a'b' + b'c'(1)
                = a'b' + b'c'
                = b'(a'+c')

Upvotes: 0

Related Questions