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To show that ( A \oplus B = (A \cup B) - (A \cap B) ), we need to prove two inclusions.

For the first inclusion, let ( x \in A \oplus B ). This means that ( x ) is in exactly one of ( A ) or ( B ), but not both. Therefore, ( x ) is in ( A ) or ( B ), but not in their intersection. Hence, ( x \in (A \cup B) - (A \cap B) ).

For the second inclusion, let ( x \in (A \cup B) - (A \cap B) ). This means that ( x ) is in either ( A ) or ( B ), but not in their intersection. Thus, ( x ) is in exactly one of ( A ) or ( B ), leading to ( x \in A \oplus B ).

Therefore, we have shown that ( A \oplus B = (A \cup B) - (A \cap B) ).

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Q: Show that a xor b a union b - a intersect b?
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Prove if a union c equals b union c and a intersect c equals b intersect c then a equals b?

suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.


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In some cases, A union B is convex, but in general this may not be true. Consider two sets A, B (subsets of Rn) such that A intersect B is the null set. Now choose a point x in A, and y in B. If a set is to be convex, then all points on the line tx + (1-t)y (0


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