Metamath Proof Explorer


Theorem 1pthd

Description: In a graph with two vertices and an edge connecting these two vertices, to go from one vertex to the other vertex via this edge is a path. The two vertices need not be distinct (in the case of a loop) - in this case, however, the path is not a simple path. (Contributed by Alexander van der Vekens, 3-Dec-2017) (Revised by AV, 22-Jan-2021) (Revised by AV, 23-Mar-2021) (Proof shortened by AV, 30-Oct-2021)

Ref Expression
Hypotheses 1wlkd.p ⊢ P = ⟨“ XY ”⟩
1wlkd.f ⊢ F = ⟨“ J ”⟩
1wlkd.x ⊢ φ → X ∈ V
1wlkd.y ⊢ φ → Y ∈ V
1wlkd.l ⊢ φ ∧ X = Y → I ⁡ J = X
1wlkd.j ⊢ φ ∧ X ≠ Y → X Y ⊆ I ⁡ J
1wlkd.v ⊢ V = Vtx ⁡ G
1wlkd.i ⊢ I = iEdg ⁡ G
Assertion 1pthd ⊢ φ → F Paths ⁡ G P

Proof

Step Hyp Ref Expression
1 1wlkd.p ⊢ P = ⟨“ XY ”⟩
2 1wlkd.f ⊢ F = ⟨“ J ”⟩
3 1wlkd.x ⊢ φ → X ∈ V
4 1wlkd.y ⊢ φ → Y ∈ V
5 1wlkd.l ⊢ φ ∧ X = Y → I ⁡ J = X
6 1wlkd.j ⊢ φ ∧ X ≠ Y → X Y ⊆ I ⁡ J
7 1wlkd.v ⊢ V = Vtx ⁡ G
8 1wlkd.i ⊢ I = iEdg ⁡ G
9 1 2 3 4 5 6 7 8 1trld ⊢ φ → F Trails ⁡ G P
10 simpr ⊢ φ ∧ F Trails ⁡ G P → F Trails ⁡ G P
11 1 2 1pthdlem1 ⊢ Fun ⁡ P ↾ 1 ..^ F -1
12 11 a1i ⊢ φ ∧ F Trails ⁡ G P → Fun ⁡ P ↾ 1 ..^ F -1
13 1 2 1pthdlem2 ⊢ P 0 F ∩ P 1 ..^ F = ∅
14 13 a1i ⊢ φ ∧ F Trails ⁡ G P → P 0 F ∩ P 1 ..^ F = ∅
15 ispth ⊢ F Paths ⁡ G P ↔ F Trails ⁡ G P ∧ Fun ⁡ P ↾ 1 ..^ F -1 ∧ P 0 F ∩ P 1 ..^ F = ∅
16 10 12 14 15 syl3anbrc ⊢ φ ∧ F Trails ⁡ G P → F Paths ⁡ G P
17 9 16 mpdan ⊢ φ → F Paths ⁡ G P