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 = ⟨“ 𝑋 𝑌 ”⟩
1wlkd.f 𝐹 = ⟨“ 𝐽 ”⟩
1wlkd.x ( 𝜑𝑋𝑉 )
1wlkd.y ( 𝜑𝑌𝑉 )
1wlkd.l ( ( 𝜑𝑋 = 𝑌 ) → ( 𝐼𝐽 ) = { 𝑋 } )
1wlkd.j ( ( 𝜑𝑋𝑌 ) → { 𝑋 , 𝑌 } ⊆ ( 𝐼𝐽 ) )
1wlkd.v 𝑉 = ( Vtx ‘ 𝐺 )
1wlkd.i 𝐼 = ( iEdg ‘ 𝐺 )
Assertion 1pthd ( 𝜑𝐹 ( Paths ‘ 𝐺 ) 𝑃 )

Proof

Step Hyp Ref Expression
1 1wlkd.p 𝑃 = ⟨“ 𝑋 𝑌 ”⟩
2 1wlkd.f 𝐹 = ⟨“ 𝐽 ”⟩
3 1wlkd.x ( 𝜑𝑋𝑉 )
4 1wlkd.y ( 𝜑𝑌𝑉 )
5 1wlkd.l ( ( 𝜑𝑋 = 𝑌 ) → ( 𝐼𝐽 ) = { 𝑋 } )
6 1wlkd.j ( ( 𝜑𝑋𝑌 ) → { 𝑋 , 𝑌 } ⊆ ( 𝐼𝐽 ) )
7 1wlkd.v 𝑉 = ( Vtx ‘ 𝐺 )
8 1wlkd.i 𝐼 = ( iEdg ‘ 𝐺 )
9 1 2 3 4 5 6 7 8 1trld ( 𝜑𝐹 ( Trails ‘ 𝐺 ) 𝑃 )
10 simpr ( ( 𝜑𝐹 ( Trails ‘ 𝐺 ) 𝑃 ) → 𝐹 ( Trails ‘ 𝐺 ) 𝑃 )
11 1 2 1pthdlem1 Fun ( 𝑃 ↾ ( 1 ..^ ( ♯ ‘ 𝐹 ) ) )
12 11 a1i ( ( 𝜑𝐹 ( Trails ‘ 𝐺 ) 𝑃 ) → Fun ( 𝑃 ↾ ( 1 ..^ ( ♯ ‘ 𝐹 ) ) ) )
13 1 2 1pthdlem2 ( ( 𝑃 “ { 0 , ( ♯ ‘ 𝐹 ) } ) ∩ ( 𝑃 “ ( 1 ..^ ( ♯ ‘ 𝐹 ) ) ) ) = ∅
14 13 a1i ( ( 𝜑𝐹 ( Trails ‘ 𝐺 ) 𝑃 ) → ( ( 𝑃 “ { 0 , ( ♯ ‘ 𝐹 ) } ) ∩ ( 𝑃 “ ( 1 ..^ ( ♯ ‘ 𝐹 ) ) ) ) = ∅ )
15 ispth ( 𝐹 ( Paths ‘ 𝐺 ) 𝑃 ↔ ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 ∧ Fun ( 𝑃 ↾ ( 1 ..^ ( ♯ ‘ 𝐹 ) ) ) ∧ ( ( 𝑃 “ { 0 , ( ♯ ‘ 𝐹 ) } ) ∩ ( 𝑃 “ ( 1 ..^ ( ♯ ‘ 𝐹 ) ) ) ) = ∅ ) )
16 10 12 14 15 syl3anbrc ( ( 𝜑𝐹 ( Trails ‘ 𝐺 ) 𝑃 ) → 𝐹 ( Paths ‘ 𝐺 ) 𝑃 )
17 9 16 mpdan ( 𝜑𝐹 ( Paths ‘ 𝐺 ) 𝑃 )