Theorem eqopab2b

Description: Equivalence of ordered pair abstraction equality and biconditional. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker eqopab2bw when possible. (Contributed by Mario Carneiro, 4-Jan-2017) (New usage is discouraged.)

	Ref	Expression
Assertion	eqopab2b	$$⊢\left\{⟨x,y⟩|\phi \right\}=\left\{⟨x,y⟩|\psi \right\}\leftrightarrow \forall x\forall y\left(\phi \leftrightarrow \psi \right)$$

Proof

Step	Hyp	Ref	Expression
1		ssopab2b	$$⊢\left\{⟨x,y⟩|\phi \right\}\subseteq \left\{⟨x,y⟩|\psi \right\}\leftrightarrow \forall x\forall y\left(\phi \to \psi \right)$$
2		ssopab2b	$$⊢\left\{⟨x,y⟩|\psi \right\}\subseteq \left\{⟨x,y⟩|\phi \right\}\leftrightarrow \forall x\forall y\left(\psi \to \phi \right)$$
3	1 2	anbi12i	$$⊢\left(\left\{⟨x,y⟩|\phi \right\}\subseteq \left\{⟨x,y⟩|\psi \right\}\wedge \left\{⟨x,y⟩|\psi \right\}\subseteq \left\{⟨x,y⟩|\phi \right\}\right)\leftrightarrow \left(\forall x\forall y\left(\phi \to \psi \right)\wedge \forall x\forall y\left(\psi \to \phi \right)\right)$$
4		eqss	$$⊢\left\{⟨x,y⟩|\phi \right\}=\left\{⟨x,y⟩|\psi \right\}\leftrightarrow \left(\left\{⟨x,y⟩|\phi \right\}\subseteq \left\{⟨x,y⟩|\psi \right\}\wedge \left\{⟨x,y⟩|\psi \right\}\subseteq \left\{⟨x,y⟩|\phi \right\}\right)$$
5		2albiim	$$⊢\forall x\forall y\left(\phi \leftrightarrow \psi \right)\leftrightarrow \left(\forall x\forall y\left(\phi \to \psi \right)\wedge \forall x\forall y\left(\psi \to \phi \right)\right)$$
6	3 4 5	3bitr4i	$$⊢\left\{⟨x,y⟩|\phi \right\}=\left\{⟨x,y⟩|\psi \right\}\leftrightarrow \forall x\forall y\left(\phi \leftrightarrow \psi \right)$$