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