Metamath Proof Explorer


Theorem ercpbllem

Description: Lemma for ercpbl . (Contributed by Mario Carneiro, 24-Feb-2015) (Revised by AV, 12-Jul-2024)

Ref Expression
Hypotheses ercpbl.r ⊢ φ → ∼ ˙ Er V
ercpbl.v ⊢ φ → V ∈ W
ercpbl.f ⊢ F = x ∈ V ⟼ x ∼ ˙
ercpbllem.1 ⊢ φ → A ∈ V
Assertion ercpbllem ⊢ φ → F ⁡ A = F ⁡ B ↔ A ∼ ˙ B

Proof

Step Hyp Ref Expression
1 ercpbl.r ⊢ φ → ∼ ˙ Er V
2 ercpbl.v ⊢ φ → V ∈ W
3 ercpbl.f ⊢ F = x ∈ V ⟼ x ∼ ˙
4 ercpbllem.1 ⊢ φ → A ∈ V
5 1 2 3 divsfval ⊢ φ → F ⁡ A = A ∼ ˙
6 1 2 3 divsfval ⊢ φ → F ⁡ B = B ∼ ˙
7 5 6 eqeq12d ⊢ φ → F ⁡ A = F ⁡ B ↔ A ∼ ˙ = B ∼ ˙
8 1 4 erth ⊢ φ → A ∼ ˙ B ↔ A ∼ ˙ = B ∼ ˙
9 7 8 bitr4d ⊢ φ → F ⁡ A = F ⁡ B ↔ A ∼ ˙ B