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Metamath Proof Explorer

Theorem eqoprab2b

Description: Equivalence of ordered pair abstraction subclass and biconditional. Compare eqopab2b . Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker eqoprab2bw when possible. (Contributed by Mario Carneiro, 4-Jan-2017) (New usage is discouraged.)

Ref Expression
Assertion eqoprab2b ⊢ x y z | φ = x y z | ψ ↔ ∀ x ∀ y ∀ z φ ↔ ψ

Proof

Step Hyp Ref Expression
1 ssoprab2b ⊢ x y z | φ ⊆ x y z | ψ ↔ ∀ x ∀ y ∀ z φ → ψ
2 ssoprab2b ⊢ x y z | ψ ⊆ x y z | φ ↔ ∀ x ∀ y ∀ z ψ → φ
3 1 2 anbi12i ⊢ x y z | φ ⊆ x y z | ψ ∧ x y z | ψ ⊆ x y z | φ ↔ ∀ x ∀ y ∀ z φ → ψ ∧ ∀ x ∀ y ∀ z ψ → φ
4 eqss ⊢ x y z | φ = x y z | ψ ↔ x y z | φ ⊆ x y z | ψ ∧ x y z | ψ ⊆ x y z | φ
5 2albiim ⊢ ∀ y ∀ z φ ↔ ψ ↔ ∀ y ∀ z φ → ψ ∧ ∀ y ∀ z ψ → φ
6 5 albii ⊢ ∀ x ∀ y ∀ z φ ↔ ψ ↔ ∀ x ∀ y ∀ z φ → ψ ∧ ∀ y ∀ z ψ → φ
7 19.26 ⊢ ∀ x ∀ y ∀ z φ → ψ ∧ ∀ y ∀ z ψ → φ ↔ ∀ x ∀ y ∀ z φ → ψ ∧ ∀ x ∀ y ∀ z ψ → φ
8 6 7 bitri ⊢ ∀ x ∀ y ∀ z φ ↔ ψ ↔ ∀ x ∀ y ∀ z φ → ψ ∧ ∀ x ∀ y ∀ z ψ → φ
9 3 4 8 3bitr4i ⊢ x y z | φ = x y z | ψ ↔ ∀ x ∀ y ∀ z φ ↔ ψ