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