Which of the following operation is commutative but not associative

To determine which gate is commutative but not associative, let's test the laws using the NAND operation (↑):

1. Commutative Law (A ↑ B = B ↑ A):

~(A . B) = ~(B . A)

Since standard multiplication (AND) is commutative, NAND is also commutative.'

2. Associative Law ((A ↑ B) ↑ C = A ↑ (B ↑ C)):

• Left-Hand Side (LHS):

  (A ↑ B) ↑ C = ~(A . B) ↑ C = ~(~( A . B) . C) = ~~(A . B) + ~C = (A . B) + ~C

• Right-Hand Side (RHS):

  A ↑ (B ↑ C) = A ↑ ~(B . C) = ~A . ~(B . C) = ~A + ~(B . C) = ~A + ~(B . C)

Since (A . B) + ~C ≠ ~ A + (B . C), the operation is not associative.