Metamath Proof Explorer


Theorem diag2f1olem

Description: Lemma for diag2f1o . (Contributed by Zhi Wang, 21-Oct-2025)

Ref Expression
Hypotheses diag2f1o.l ⊢ L = C Δ func D
diag2f1o.a ⊢ A = Base C
diag2f1o.h ⊢ H = Hom ⁡ C
diag2f1o.x ⊢ φ → X ∈ A
diag2f1o.y ⊢ φ → Y ∈ A
diag2f1o.n ⊢ N = D Nat C
diag2f1o.d No typesetting found for |- ( ph -> D e. TermCat ) with typecode |-
diag2f1olem.m ⊢ φ → M ∈ 1 st ⁡ L ⁡ X N 1 st ⁡ L ⁡ Y
diag2f1olem.b ⊢ B = Base D
diag2f1olem.z ⊢ φ → Z ∈ B
diag2f1olem.f ⊢ F = M ⁡ Z
Assertion diag2f1olem ⊢ φ → F ∈ X H Y ∧ M = X 2 nd ⁡ L Y ⁡ F

Proof

Step Hyp Ref Expression
1 diag2f1o.l ⊢ L = C Δ func D
2 diag2f1o.a ⊢ A = Base C
3 diag2f1o.h ⊢ H = Hom ⁡ C
4 diag2f1o.x ⊢ φ → X ∈ A
5 diag2f1o.y ⊢ φ → Y ∈ A
6 diag2f1o.n ⊢ N = D Nat C
7 diag2f1o.d Could not format ( ph -> D e. TermCat ) : No typesetting found for |- ( ph -> D e. TermCat ) with typecode |-
8 diag2f1olem.m ⊢ φ → M ∈ 1 st ⁡ L ⁡ X N 1 st ⁡ L ⁡ Y
9 diag2f1olem.b ⊢ B = Base D
10 diag2f1olem.z ⊢ φ → Z ∈ B
11 diag2f1olem.f ⊢ F = M ⁡ Z
12 6 8 nat1st2nd ⊢ φ → M ∈ 1 st ⁡ 1 st ⁡ L ⁡ X 2 nd ⁡ 1 st ⁡ L ⁡ X N 1 st ⁡ 1 st ⁡ L ⁡ Y 2 nd ⁡ 1 st ⁡ L ⁡ Y
13 6 12 9 3 10 natcl ⊢ φ → M ⁡ Z ∈ 1 st ⁡ 1 st ⁡ L ⁡ X ⁡ Z H 1 st ⁡ 1 st ⁡ L ⁡ Y ⁡ Z
14 6 12 natrcl2 ⊢ φ → 1 st ⁡ 1 st ⁡ L ⁡ X D Func C 2 nd ⁡ 1 st ⁡ L ⁡ X
15 14 funcrcl3 ⊢ φ → C ∈ Cat
16 7 termccatd ⊢ φ → D ∈ Cat
17 eqid ⊢ 1 st ⁡ L ⁡ X = 1 st ⁡ L ⁡ X
18 1 15 16 2 4 17 9 10 diag11 ⊢ φ → 1 st ⁡ 1 st ⁡ L ⁡ X ⁡ Z = X
19 eqid ⊢ 1 st ⁡ L ⁡ Y = 1 st ⁡ L ⁡ Y
20 1 15 16 2 5 19 9 10 diag11 ⊢ φ → 1 st ⁡ 1 st ⁡ L ⁡ Y ⁡ Z = Y
21 18 20 oveq12d ⊢ φ → 1 st ⁡ 1 st ⁡ L ⁡ X ⁡ Z H 1 st ⁡ 1 st ⁡ L ⁡ Y ⁡ Z = X H Y
22 13 21 eleqtrd ⊢ φ → M ⁡ Z ∈ X H Y
23 11 22 eqeltrid ⊢ φ → F ∈ X H Y
24 7 6 8 9 10 11 termcnatval ⊢ φ → M = Z F
25 1 2 9 3 15 16 4 5 23 diag2 ⊢ φ → X 2 nd ⁡ L Y ⁡ F = B × F
26 7 9 10 termcbas2 ⊢ φ → B = Z
27 26 xpeq1d ⊢ φ → B × F = Z × F
28 xpsng ⊢ Z ∈ B ∧ F ∈ X H Y → Z × F = Z F
29 10 23 28 syl2anc ⊢ φ → Z × F = Z F
30 25 27 29 3eqtrd ⊢ φ → X 2 nd ⁡ L Y ⁡ F = Z F
31 24 30 eqtr4d ⊢ φ → M = X 2 nd ⁡ L Y ⁡ F
32 23 31 jca ⊢ φ → F ∈ X H Y ∧ M = X 2 nd ⁡ L Y ⁡ F