Theorem eqss

Description: The subclass relationship is antisymmetric. Compare Theorem 4 of Suppes p. 22. (Contributed by NM, 21-May-1993)

	Ref	Expression
Assertion	eqss	$$⊢A=B\leftrightarrow \left(A\subseteq B\wedge B\subseteq A\right)$$

Proof

Step	Hyp	Ref	Expression
1		albiim	$$⊢\forall x\left(x\in A\leftrightarrow x\in B\right)\leftrightarrow \left(\forall x\left(x\in A\to x\in B\right)\wedge \forall x\left(x\in B\to x\in A\right)\right)$$
2		dfcleq	$$⊢A=B\leftrightarrow \forall x\left(x\in A\leftrightarrow x\in B\right)$$
3		dfss2	$$⊢A\subseteq B\leftrightarrow \forall x\left(x\in A\to x\in B\right)$$
4		dfss2	$$⊢B\subseteq A\leftrightarrow \forall x\left(x\in B\to x\in A\right)$$
5	3 4	anbi12i	$$⊢\left(A\subseteq B\wedge B\subseteq A\right)\leftrightarrow \left(\forall x\left(x\in A\to x\in B\right)\wedge \forall x\left(x\in B\to x\in A\right)\right)$$
6	1 2 5	3bitr4i	$$⊢A=B\leftrightarrow \left(A\subseteq B\wedge B\subseteq A\right)$$