Question

represent the simplest reduction of

(ab' + ac)'

a) a' + bc'

b) a' + b'c

c) a + bc'

d) a + bc

Answer 1

Given function is : (ab' + ac)'

=> (ab' + ac)'
=>(ab')' (ac)' (A+B)' = A'.B' - de Morgan’s Theorem

=>(a' + (b')') (a' + c') (A.B)' = A'+ B' - de Morgan’s Theorem
=>(a'+b) (a'+c') (A')' = A Double Negation Law
=> a'a' + ba'+a'c'+ bc'
=> a' + ba' + a'c' + bc' A.A = A   Idempotent Law
=> a'(1+b) + a'c' + bc' A + 1 = 1 Annulment Law
=> a'+ a'c' + bc'
=>a'(1+c') + bc' A + 1 = 1 Annulment Law
=> a'+ bc'

So anwer is : a) a'+ bc'