Metamath Proof Explorer


Theorem uhgrwkspthlem2

Description: Lemma 2 for uhgrwkspth . (Contributed by AV, 25-Jan-2021)

Ref Expression
Assertion uhgrwkspthlem2 ⊢ F Walks ⁡ G P ∧ F = 1 ∧ A ≠ B ∧ P ⁡ 0 = A ∧ P ⁡ F = B → Fun ⁡ P -1

Proof

Step Hyp Ref Expression
1 eqid ⊢ Vtx ⁡ G = Vtx ⁡ G
2 1 wlkp ⊢ F Walks ⁡ G P → P : 0 … F ⟶ Vtx ⁡ G
3 oveq2 ⊢ F = 1 → 0 … F = 0 … 1
4 1e0p1 ⊢ 1 = 0 + 1
5 4 oveq2i ⊢ 0 … 1 = 0 … 0 + 1
6 0z ⊢ 0 ∈ ℤ
7 fzpr ⊢ 0 ∈ ℤ → 0 … 0 + 1 = 0 0 + 1
8 6 7 ax-mp ⊢ 0 … 0 + 1 = 0 0 + 1
9 0p1e1 ⊢ 0 + 1 = 1
10 9 preq2i ⊢ 0 0 + 1 = 0 1
11 5 8 10 3eqtri ⊢ 0 … 1 = 0 1
12 3 11 eqtrdi ⊢ F = 1 → 0 … F = 0 1
13 12 feq2d ⊢ F = 1 → P : 0 … F ⟶ Vtx ⁡ G ↔ P : 0 1 ⟶ Vtx ⁡ G
14 13 adantr ⊢ F = 1 ∧ P ⁡ 0 = A ∧ P ⁡ F = B → P : 0 … F ⟶ Vtx ⁡ G ↔ P : 0 1 ⟶ Vtx ⁡ G
15 simpl ⊢ P ⁡ 0 = A ∧ P ⁡ F = B → P ⁡ 0 = A
16 simpr ⊢ P ⁡ 0 = A ∧ P ⁡ F = B → P ⁡ F = B
17 15 16 neeq12d ⊢ P ⁡ 0 = A ∧ P ⁡ F = B → P ⁡ 0 ≠ P ⁡ F ↔ A ≠ B
18 17 bicomd ⊢ P ⁡ 0 = A ∧ P ⁡ F = B → A ≠ B ↔ P ⁡ 0 ≠ P ⁡ F
19 fveq2 ⊢ F = 1 → P ⁡ F = P ⁡ 1
20 19 neeq2d ⊢ F = 1 → P ⁡ 0 ≠ P ⁡ F ↔ P ⁡ 0 ≠ P ⁡ 1
21 18 20 sylan9bbr ⊢ F = 1 ∧ P ⁡ 0 = A ∧ P ⁡ F = B → A ≠ B ↔ P ⁡ 0 ≠ P ⁡ 1
22 14 21 anbi12d ⊢ F = 1 ∧ P ⁡ 0 = A ∧ P ⁡ F = B → P : 0 … F ⟶ Vtx ⁡ G ∧ A ≠ B ↔ P : 0 1 ⟶ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1
23 1z ⊢ 1 ∈ ℤ
24 fpr2g ⊢ 0 ∈ ℤ ∧ 1 ∈ ℤ → P : 0 1 ⟶ Vtx ⁡ G ↔ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P = 0 P ⁡ 0 1 P ⁡ 1
25 6 23 24 mp2an ⊢ P : 0 1 ⟶ Vtx ⁡ G ↔ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P = 0 P ⁡ 0 1 P ⁡ 1
26 funcnvs2 ⊢ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ ⟨“ P ⁡ 0 P ⁡ 1 ”⟩ -1
27 26 3expa ⊢ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ ⟨“ P ⁡ 0 P ⁡ 1 ”⟩ -1
28 27 adantl ⊢ P = 0 P ⁡ 0 1 P ⁡ 1 ∧ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ ⟨“ P ⁡ 0 P ⁡ 1 ”⟩ -1
29 simpl ⊢ P = 0 P ⁡ 0 1 P ⁡ 1 ∧ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → P = 0 P ⁡ 0 1 P ⁡ 1
30 s2prop ⊢ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G → ⟨“ P ⁡ 0 P ⁡ 1 ”⟩ = 0 P ⁡ 0 1 P ⁡ 1
31 30 eqcomd ⊢ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G → 0 P ⁡ 0 1 P ⁡ 1 = ⟨“ P ⁡ 0 P ⁡ 1 ”⟩
32 31 adantr ⊢ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → 0 P ⁡ 0 1 P ⁡ 1 = ⟨“ P ⁡ 0 P ⁡ 1 ”⟩
33 32 adantl ⊢ P = 0 P ⁡ 0 1 P ⁡ 1 ∧ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → 0 P ⁡ 0 1 P ⁡ 1 = ⟨“ P ⁡ 0 P ⁡ 1 ”⟩
34 29 33 eqtrd ⊢ P = 0 P ⁡ 0 1 P ⁡ 1 ∧ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → P = ⟨“ P ⁡ 0 P ⁡ 1 ”⟩
35 34 cnveqd ⊢ P = 0 P ⁡ 0 1 P ⁡ 1 ∧ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → P -1 = ⟨“ P ⁡ 0 P ⁡ 1 ”⟩ -1
36 35 funeqd ⊢ P = 0 P ⁡ 0 1 P ⁡ 1 ∧ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ P -1 ↔ Fun ⁡ ⟨“ P ⁡ 0 P ⁡ 1 ”⟩ -1
37 28 36 mpbird ⊢ P = 0 P ⁡ 0 1 P ⁡ 1 ∧ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ P -1
38 37 exp32 ⊢ P = 0 P ⁡ 0 1 P ⁡ 1 → P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G → P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ P -1
39 38 impcom ⊢ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P = 0 P ⁡ 0 1 P ⁡ 1 → P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ P -1
40 39 3impa ⊢ P ⁡ 0 ∈ Vtx ⁡ G ∧ P ⁡ 1 ∈ Vtx ⁡ G ∧ P = 0 P ⁡ 0 1 P ⁡ 1 → P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ P -1
41 25 40 sylbi ⊢ P : 0 1 ⟶ Vtx ⁡ G → P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ P -1
42 41 imp ⊢ P : 0 1 ⟶ Vtx ⁡ G ∧ P ⁡ 0 ≠ P ⁡ 1 → Fun ⁡ P -1
43 22 42 biimtrdi ⊢ F = 1 ∧ P ⁡ 0 = A ∧ P ⁡ F = B → P : 0 … F ⟶ Vtx ⁡ G ∧ A ≠ B → Fun ⁡ P -1
44 43 expd ⊢ F = 1 ∧ P ⁡ 0 = A ∧ P ⁡ F = B → P : 0 … F ⟶ Vtx ⁡ G → A ≠ B → Fun ⁡ P -1
45 44 com12 ⊢ P : 0 … F ⟶ Vtx ⁡ G → F = 1 ∧ P ⁡ 0 = A ∧ P ⁡ F = B → A ≠ B → Fun ⁡ P -1
46 45 expd ⊢ P : 0 … F ⟶ Vtx ⁡ G → F = 1 → P ⁡ 0 = A ∧ P ⁡ F = B → A ≠ B → Fun ⁡ P -1
47 46 com34 ⊢ P : 0 … F ⟶ Vtx ⁡ G → F = 1 → A ≠ B → P ⁡ 0 = A ∧ P ⁡ F = B → Fun ⁡ P -1
48 47 impd ⊢ P : 0 … F ⟶ Vtx ⁡ G → F = 1 ∧ A ≠ B → P ⁡ 0 = A ∧ P ⁡ F = B → Fun ⁡ P -1
49 2 48 syl ⊢ F Walks ⁡ G P → F = 1 ∧ A ≠ B → P ⁡ 0 = A ∧ P ⁡ F = B → Fun ⁡ P -1
50 49 3imp ⊢ F Walks ⁡ G P ∧ F = 1 ∧ A ≠ B ∧ P ⁡ 0 = A ∧ P ⁡ F = B → Fun ⁡ P -1