prove that for sets SS, AA, BB, and CC, where AA, BB, and CC are subsets of SS, the following equality holds: (A−B)−C=(A−C)−(B−C)



Answer to a math question prove that for sets SS, AA, BB, and CC, where AA, BB, and CC are subsets of SS, the following equality holds: (A−B)−C=(A−C)−(B−C)

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To prove the given equality (A - B) - C = (A - C) - (B - C), we need to show that both sides are subsets of each other.

Let x be an arbitrary element in (A - B) - C.

This means that x is in (A - B) and not in C.

To be in (A - B), x must be in A and not in B.

So, x is in A and not in B, and x is not in C.

Now, let's consider (A - C) - (B - C).

Let y be an arbitrary element in (A - C) - (B - C).

This means that y is in (A - C) and not in (B - C).

To be in (A - C), y must be in A and not in C.

To not be in (B - C), y must either not be in B or be in C.

Since y is not in C, it follows that y is not in B.

Therefore, y is in A and not in B, and y is not in C.

Thus, (A - B) - C is a subset of (A - C) - (B - C).

To show the reverse inclusion, we can follow a similar argument.

Let z be an arbitrary element in (A - C) - (B - C).

This means that z is in A and not in C, and z is not in B or is in C.

Since z is not in B or is in C, it follows that z is not in B.

Therefore, z is in (A - B) and not in C.

Hence, (A - C) - (B - C) is a subset of (A - B) - C.

Since both sides of the equality are subsets of each other, we can conclude that (A - B) - C = (A - C) - (B - C).

Answer: (A - B) - C = (A - C) - (B - C)

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