Verify [itex]A\cap(B-C)=(A\cap B)-(A\cap C)[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

This is how I did it:

From [itex]x\in[A\cap(B-C)][/itex] (lhs) we have that

1.) [itex]x\in A[/itex]

2.) [itex]x\in B[/itex]

3.) [itex]x\notin C[/itex]

From #1 and #2 we have that

4.) [itex]x\in(A\cap B)[/itex]

Now, since [itex]x\in(A\cap C)[/itex] means that [itex]x\in A[/itex] and [itex]x\in C[/itex], but we have [itex]x\notin C[/itex], it follows that

5.) [itex]x\notin(A\cap C)[/itex]

Remembering that [itex]x\in(A-B)[/itex] means that [itex]x\in A[/itex] and [itex]x\notin B[/itex] it follows from #4 and #5 that

[itex]x\in[(A\cap B)-(A\cap C)][/itex] (rhs)

Is this right?

- Kamataat

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