1. ## Is my set thoery proof correct?

Prove or disprove: For all sets A and B, (A-B) U (A intersect B) = A

My attempt at a proof:
This statement is true.

Suppose that A and B are sets where (A-B) U (A intersect B) and x is an element (A-B) U (A intersect B).

By definition of Union x is an element A-B OR x is an element of A intersect B (but not both).

So if x is an element of (A-B) then by definition of difference x is an element of A and x is NOT an element of B. Therefore, A-B=A.

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If x is an element of A intersection B then x is an element of A and x is an element of B by definition of intersection.

Therefore, x is an element of A intersect B.

So, if x is an element of A intersect B then x is NOT and element of A-B and if x is an element of A-B then x is NOT an element of A intersect B.

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A-B=A because by definition of difference if x is an element of A-B then x is an element of A but it is NOT an element of B. So, x is an element of A which equals A.

If that is NOT the case, then,

x is must be an element of A intersect B, which means it is an element of A and an element of B, so either way if x is an element of B or not an element of B it is always an element of A and a set is always a subset of itself. Therefore, the original statement IS true and for all sets A and B, (A-B) U (A intersect B) = A
END OF PROOF.

2. Originally Posted by matthayzon89
Prove or disprove: For all sets A and B, (A-B) U (A intersect B) = A

My attempt at a proof:
This statement is true.

Suppose that A and B are sets where (A-B) U (A intersect B) and x is an element (A-B) U (A intersect B).

By definition of Union x is an element A-B OR x is an element of A intersect B (but not both).

So if x is an element of (A-B) then by definition of difference x is an element of A and x is NOT an element of B. Therefore, A-B=A.

___________________________________

If x is an element of A intersection B then x is an element of A and x is an element of B by definition of intersection.

Therefore, x is an element of A intersect B.

So, if x is an element of A intersect B then x is NOT and element of A-B and if x is an element of A-B then x is NOT an element of A intersect B.

__________________________________________
A-B=A because by definition of difference if x is an element of A-B then x is an element of A but it is NOT an element of B. So, x is an element of A which equals A.

If that is NOT the case, then,

x is must be an element of A intersect B, which means it is an element of A and an element of B, so either way if x is an element of B or not an element of B it is always an element of A and a set is always a subset of itself. Therefore, the original statement IS true and for all sets A and B, (A-B) U (A intersect B) = A
END OF PROOF.

This is too long......and at least twice I saw $A-B=A$ which, of course, is false in general.

1) Suppose $x\in (A-B)\cup(A\cap B)\Longrightarrow x\in A-B\,\,\,or\,\,\,x\in A\cap B\Longrightarrow \,\,anyway\,,\,\,x\in A$ $\Longrightarrow (A-B)\cup(A\cap B)\subset A$

2) Suppose now $x\in A$ ; if also $x\in B\,\,\,then\,\,\,x\in A\cap B$ , otherwise $x\notin B\Longrightarrow x\in A-B\Longrightarrow \,\,\,anyway\,\,\,x\in (A-B)\cup(A\cap B)\Longrightarrow A\subset (A-B)\cup(A\cap B)$

Tonio