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I need to prove this: If $A \subseteq B$ and $B\subseteq A$, then $A = B$.

I know that $A \subseteq B \implies (x \in A\implies x \in B)$ and similarly $B \subseteq A \implies (x \in B \implies x \in A)$.

How can I prove that $A=B$?

So, the result is:

$$(x\notin A \wedge x \in A)\vee (x\notin A \wedge x\notin B)\vee (x\in B\wedge x\notin B)\vee (x\in B\wedge x\in A)$$

But plainly $x \notin A \wedge x \in A$ and $x \in B \wedge x \notin B$ are false, so

$$(x \notin A \wedge x \notin B) \vee (x \in A \wedge x \in B)$$

Is that correct?

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up vote 3 down vote accepted

$A\subseteq B$ means that $x\in A\implies x\in B$. $B\subseteq A$ means that $x\in B\implies x\in A$. But $A=B$ means that $x\in A\iff x\in B$. Therefore, $A\subseteq B$ and $B\subseteq A$ if and only if $A=B$.

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so, i can use biconditional to proof using logic? x<=>y is what i did, right? – Matheus Silva Aug 11 '13 at 22:15
i think did not express myself very well, so, what i want, is to proof that A=B using A⊆B and B⊆A – Matheus Silva Aug 11 '13 at 22:22
is that u post? if was, that SURE straightfoward...i had learn some different way...thats why i'm talking about roundabout... – Matheus Silva Aug 11 '13 at 22:29

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