Theorem opabbi

Description: Equality deduction for class abstraction of ordered pairs. (Contributed by Giovanni Mascellani, 10-Apr-2018)

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

Proof

Step	Hyp	Ref	Expression
1		eqopab2b	$$⊢\left\{⟨x,y⟩|\phi \right\}=\left\{⟨x,y⟩|\psi \right\}\leftrightarrow \forall x\forall y\left(\phi \leftrightarrow \psi \right)$$
2	1	biimpri	$$⊢\forall x\forall y\left(\phi \leftrightarrow \psi \right)\to \left\{⟨x,y⟩|\phi \right\}=\left\{⟨x,y⟩|\psi \right\}$$