Theorem oprabbi

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

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

Proof

Step	Hyp	Ref	Expression
1		eqoprab2b	$$⊢\left\{⟨⟨x,y⟩,z⟩|\phi \right\}=\left\{⟨⟨x,y⟩,z⟩|\psi \right\}\leftrightarrow \forall x\forall y\forall z\left(\phi \leftrightarrow \psi \right)$$
2	1	biimpri	$$⊢\forall x\forall y\forall z\left(\phi \leftrightarrow \psi \right)\to \left\{⟨⟨x,y⟩,z⟩|\phi \right\}=\left\{⟨⟨x,y⟩,z⟩|\psi \right\}$$