Metamath Proof Explorer


Theorem diag2f1lem

Description: Lemma for diag2f1 . The converse is trivial ( fveq2 ). (Contributed by Zhi Wang, 21-Oct-2025)

Ref Expression
Hypotheses diag2f1.l ⊢ L = C Δ func D
diag2f1.a ⊢ A = Base C
diag2f1.b ⊢ B = Base D
diag2f1.h ⊢ H = Hom ⁡ C
diag2f1.c ⊢ φ → C ∈ Cat
diag2f1.d ⊢ φ → D ∈ Cat
diag2f1.x ⊢ φ → X ∈ A
diag2f1.y ⊢ φ → Y ∈ A
diag2f1.0 ⊢ φ → B ≠ ∅
diag2f1lem.f ⊢ φ → F ∈ X H Y
diag2f1lem.g ⊢ φ → G ∈ X H Y
Assertion diag2f1lem ⊢ φ → X 2 nd ⁡ L Y ⁡ F = X 2 nd ⁡ L Y ⁡ G → F = G

Proof

Step Hyp Ref Expression
1 diag2f1.l ⊢ L = C Δ func D
2 diag2f1.a ⊢ A = Base C
3 diag2f1.b ⊢ B = Base D
4 diag2f1.h ⊢ H = Hom ⁡ C
5 diag2f1.c ⊢ φ → C ∈ Cat
6 diag2f1.d ⊢ φ → D ∈ Cat
7 diag2f1.x ⊢ φ → X ∈ A
8 diag2f1.y ⊢ φ → Y ∈ A
9 diag2f1.0 ⊢ φ → B ≠ ∅
10 diag2f1lem.f ⊢ φ → F ∈ X H Y
11 diag2f1lem.g ⊢ φ → G ∈ X H Y
12 1 2 3 4 5 6 7 8 10 diag2 ⊢ φ → X 2 nd ⁡ L Y ⁡ F = B × F
13 1 2 3 4 5 6 7 8 11 diag2 ⊢ φ → X 2 nd ⁡ L Y ⁡ G = B × G
14 12 13 eqeq12d ⊢ φ → X 2 nd ⁡ L Y ⁡ F = X 2 nd ⁡ L Y ⁡ G ↔ B × F = B × G
15 xpcan ⊢ B ≠ ∅ → B × F = B × G ↔ F = G
16 9 15 syl ⊢ φ → B × F = B × G ↔ F = G
17 14 16 bitrd ⊢ φ → X 2 nd ⁡ L Y ⁡ F = X 2 nd ⁡ L Y ⁡ G ↔ F = G
18 sneqrg ⊢ F ∈ X H Y → F = G → F = G
19 10 18 syl ⊢ φ → F = G → F = G
20 17 19 sylbid ⊢ φ → X 2 nd ⁡ L Y ⁡ F = X 2 nd ⁡ L Y ⁡ G → F = G