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 ) )