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# 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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## 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=$A−C$−$B−C$

Miles
4.9
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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