(a) x⊕y=(xy+x′y′)′ =(xy)′ =x⊕y, it is valid. (x+y)⊕z=(x+y)z+(x+y)z (b)=xyz+xz+yz 1 4,6 2,6 =Σm(1,2,4,6) x⊕(y+z)=x(y+z)+x(y+z) =xy+xz+yz 2,31,34 =∑m(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=(x+y)(x+y) =(x+y)xy =(x+y)0 =(x+y), so it is valid.