state associative law. prove or disprove algebraically if associative law holds for NOR operation or not.

To prove or disprove the associative law for the NOR operation algebraically, we need to show whether the expression (A NOR B) NOR C is equal to A NOR (B NOR C) for all possible values of A, B, and C.

Let's start by defining the NOR operation. NOR is a logical operation that returns true (1) if both inputs are false (0), and false (0) otherwise. The truth table for NOR is as follows:

A | B | A NOR B

0 | 0 | 1 0 | 1 | 0 1 | 0 | 0 1 | 1 | 0

Now, let's evaluate the expression (A NOR B) NOR C:

(A NOR B) NOR C = (A NOR B) NOR C = ((A NOR B) NOR C) NOR ((A NOR B) NOR C)

Next, let's evaluate the expression A NOR (B NOR C):

A NOR (B NOR C) = A NOR (B NOR C) = (A NOR (B NOR C)) NOR (A NOR (B NOR C))

To prove the associative law, we need to show that (A NOR B) NOR C is equal to A NOR (B NOR C) for all possible values of A, B, and C. However, by comparing the two expressions above, we can see that they are not equal.

For example, let's consider the case where A = 0, B = 1, and C = 0:

(A NOR B) NOR C = (0 NOR 1) NOR 0 = 0 NOR 0 = 1

A NOR (B NOR C) = 0 NOR (1 NOR 0) = 0 NOR 1 = 0

Since (A NOR B) NOR C is not equal to A NOR (B NOR C) for this particular case, we can conclude that the associative law does not hold for the NOR operation algebraically.

Therefore, we have disproven the associative law for the NOR operation algebraically.