Explanation
(a) x ⊕ y = (xy + x′y ′)′
= (xy )′
= x ⊕ y, it is valid.
(b)
= Σm(1, 2, 4, 6)
= Σm(1, 2, 3, 4)
(x + y) ⊕ z ≠ x ⊕ (y + z)
So option (b) is invalid.
(c) (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)
Associativity is true on Ex-OR operator so it valid.
(d)
= (x + y), so it is valid.