Metamath Proof Explorer


Theorem f1otrgds

Description: Convenient lemma for f1otrg . (Contributed by Thierry Arnoux, 19-Mar-2019)

Ref Expression
Hypotheses f1otrkg.p ⊢ P = Base G
f1otrkg.d ⊢ D = dist ⁡ G
f1otrkg.i ⊢ I = Itv ⁡ G
f1otrkg.b ⊢ B = Base H
f1otrkg.e ⊢ E = dist ⁡ H
f1otrkg.j ⊢ J = Itv ⁡ H
f1otrkg.f ⊢ φ → F : B ⟶ 1-1 onto P
f1otrkg.1 ⊢ φ ∧ e ∈ B ∧ f ∈ B → e E f = F ⁡ e D F ⁡ f
f1otrkg.2 ⊢ φ ∧ e ∈ B ∧ f ∈ B ∧ g ∈ B → g ∈ e J f ↔ F ⁡ g ∈ F ⁡ e I F ⁡ f
f1otrgitv.x ⊢ φ → X ∈ B
f1otrgitv.y ⊢ φ → Y ∈ B
Assertion f1otrgds ⊢ φ → X E Y = F ⁡ X D F ⁡ Y

Proof

Step Hyp Ref Expression
1 f1otrkg.p ⊢ P = Base G
2 f1otrkg.d ⊢ D = dist ⁡ G
3 f1otrkg.i ⊢ I = Itv ⁡ G
4 f1otrkg.b ⊢ B = Base H
5 f1otrkg.e ⊢ E = dist ⁡ H
6 f1otrkg.j ⊢ J = Itv ⁡ H
7 f1otrkg.f ⊢ φ → F : B ⟶ 1-1 onto P
8 f1otrkg.1 ⊢ φ ∧ e ∈ B ∧ f ∈ B → e E f = F ⁡ e D F ⁡ f
9 f1otrkg.2 ⊢ φ ∧ e ∈ B ∧ f ∈ B ∧ g ∈ B → g ∈ e J f ↔ F ⁡ g ∈ F ⁡ e I F ⁡ f
10 f1otrgitv.x ⊢ φ → X ∈ B
11 f1otrgitv.y ⊢ φ → Y ∈ B
12 8 ralrimivva ⊢ φ → ∀ e ∈ B ∀ f ∈ B e E f = F ⁡ e D F ⁡ f
13 oveq1 ⊢ e = X → e E f = X E f
14 fveq2 ⊢ e = X → F ⁡ e = F ⁡ X
15 14 oveq1d ⊢ e = X → F ⁡ e D F ⁡ f = F ⁡ X D F ⁡ f
16 13 15 eqeq12d ⊢ e = X → e E f = F ⁡ e D F ⁡ f ↔ X E f = F ⁡ X D F ⁡ f
17 oveq2 ⊢ f = Y → X E f = X E Y
18 fveq2 ⊢ f = Y → F ⁡ f = F ⁡ Y
19 18 oveq2d ⊢ f = Y → F ⁡ X D F ⁡ f = F ⁡ X D F ⁡ Y
20 17 19 eqeq12d ⊢ f = Y → X E f = F ⁡ X D F ⁡ f ↔ X E Y = F ⁡ X D F ⁡ Y
21 16 20 rspc2v ⊢ X ∈ B ∧ Y ∈ B → ∀ e ∈ B ∀ f ∈ B e E f = F ⁡ e D F ⁡ f → X E Y = F ⁡ X D F ⁡ Y
22 10 11 21 syl2anc ⊢ φ → ∀ e ∈ B ∀ f ∈ B e E f = F ⁡ e D F ⁡ f → X E Y = F ⁡ X D F ⁡ Y
23 12 22 mpd ⊢ φ → X E Y = F ⁡ X D F ⁡ Y