Metamath Proof Explorer


Theorem 2trld

Description: Construction of a trail from two given edges in a graph. (Contributed by Alexander van der Vekens, 4-Dec-2017) (Revised by AV, 24-Jan-2021) (Revised by AV, 24-Mar-2021) (Proof shortened by AV, 30-Oct-2021)

Ref Expression
Hypotheses 2wlkd.p ⊢ P = ⟨“ ABC ”⟩
2wlkd.f ⊢ F = ⟨“ JK ”⟩
2wlkd.s ⊢ φ → A ∈ V ∧ B ∈ V ∧ C ∈ V
2wlkd.n ⊢ φ → A ≠ B ∧ B ≠ C
2wlkd.e ⊢ φ → A B ⊆ I ⁡ J ∧ B C ⊆ I ⁡ K
2wlkd.v ⊢ V = Vtx ⁡ G
2wlkd.i ⊢ I = iEdg ⁡ G
2trld.n ⊢ φ → J ≠ K
Assertion 2trld ⊢ φ → F Trails ⁡ G P

Proof

Step Hyp Ref Expression
1 2wlkd.p ⊢ P = ⟨“ ABC ”⟩
2 2wlkd.f ⊢ F = ⟨“ JK ”⟩
3 2wlkd.s ⊢ φ → A ∈ V ∧ B ∈ V ∧ C ∈ V
4 2wlkd.n ⊢ φ → A ≠ B ∧ B ≠ C
5 2wlkd.e ⊢ φ → A B ⊆ I ⁡ J ∧ B C ⊆ I ⁡ K
6 2wlkd.v ⊢ V = Vtx ⁡ G
7 2wlkd.i ⊢ I = iEdg ⁡ G
8 2trld.n ⊢ φ → J ≠ K
9 1 2 3 4 5 6 7 2wlkd ⊢ φ → F Walks ⁡ G P
10 1 2 3 4 5 2wlkdlem7 ⊢ φ → J ∈ V ∧ K ∈ V
11 df-3an ⊢ J ∈ V ∧ K ∈ V ∧ J ≠ K ↔ J ∈ V ∧ K ∈ V ∧ J ≠ K
12 10 8 11 sylanbrc ⊢ φ → J ∈ V ∧ K ∈ V ∧ J ≠ K
13 funcnvs2 ⊢ J ∈ V ∧ K ∈ V ∧ J ≠ K → Fun ⁡ ⟨“ JK ”⟩ -1
14 12 13 syl ⊢ φ → Fun ⁡ ⟨“ JK ”⟩ -1
15 2 cnveqi ⊢ F -1 = ⟨“ JK ”⟩ -1
16 15 funeqi ⊢ Fun ⁡ F -1 ↔ Fun ⁡ ⟨“ JK ”⟩ -1
17 14 16 sylibr ⊢ φ → Fun ⁡ F -1
18 istrl ⊢ F Trails ⁡ G P ↔ F Walks ⁡ G P ∧ Fun ⁡ F -1
19 9 17 18 sylanbrc ⊢ φ → F Trails ⁡ G P