Metamath Proof Explorer


Theorem scottab

Description: Value of the Scott operation at a class abstraction. (Contributed by Rohan Ridenour, 14-Aug-2023)

Ref Expression
Hypothesis scottab.1 ⊢ x = y → φ ↔ ψ
Assertion scottab ⊢ Scott x | φ = x | φ ∧ ∀ y ψ → rank ⁡ x ⊆ rank ⁡ y

Proof

Step Hyp Ref Expression
1 scottab.1 ⊢ x = y → φ ↔ ψ
2 nfv ⊢ Ⅎ x ψ
3 2 1 scottabf ⊢ Scott x | φ = x | φ ∧ ∀ y ψ → rank ⁡ x ⊆ rank ⁡ y