Metamath Proof Explorer


Theorem spthispth

Description: A simple path is a path (in an undirected graph). (Contributed by Alexander van der Vekens, 21-Oct-2017) (Revised by AV, 9-Jan-2021) (Proof shortened by AV, 30-Oct-2021)

Ref Expression
Assertion spthispth ⊢ F SPaths ⁡ G P → F Paths ⁡ G P

Proof

Step Hyp Ref Expression
1 simpl ⊢ F Trails ⁡ G P ∧ Fun ⁡ P -1 → F Trails ⁡ G P
2 funres11 ⊢ Fun ⁡ P -1 → Fun ⁡ P ↾ 1 ..^ F -1
3 2 adantl ⊢ F Trails ⁡ G P ∧ Fun ⁡ P -1 → Fun ⁡ P ↾ 1 ..^ F -1
4 imain ⊢ Fun ⁡ P -1 → P 0 F ∩ 1 ..^ F = P 0 F ∩ P 1 ..^ F
5 1e0p1 ⊢ 1 = 0 + 1
6 5 oveq1i ⊢ 1 ..^ F = 0 + 1 ..^ F
7 6 ineq2i ⊢ 0 F ∩ 1 ..^ F = 0 F ∩ 0 + 1 ..^ F
8 0z ⊢ 0 ∈ ℤ
9 prinfzo0 ⊢ 0 ∈ ℤ → 0 F ∩ 0 + 1 ..^ F = ∅
10 8 9 ax-mp ⊢ 0 F ∩ 0 + 1 ..^ F = ∅
11 7 10 eqtri ⊢ 0 F ∩ 1 ..^ F = ∅
12 11 imaeq2i ⊢ P 0 F ∩ 1 ..^ F = P ∅
13 ima0 ⊢ P ∅ = ∅
14 12 13 eqtri ⊢ P 0 F ∩ 1 ..^ F = ∅
15 4 14 eqtr3di ⊢ Fun ⁡ P -1 → P 0 F ∩ P 1 ..^ F = ∅
16 15 adantl ⊢ F Trails ⁡ G P ∧ Fun ⁡ P -1 → P 0 F ∩ P 1 ..^ F = ∅
17 1 3 16 3jca ⊢ F Trails ⁡ G P ∧ Fun ⁡ P -1 → F Trails ⁡ G P ∧ Fun ⁡ P ↾ 1 ..^ F -1 ∧ P 0 F ∩ P 1 ..^ F = ∅
18 isspth ⊢ F SPaths ⁡ G P ↔ F Trails ⁡ G P ∧ Fun ⁡ P -1
19 ispth ⊢ F Paths ⁡ G P ↔ F Trails ⁡ G P ∧ Fun ⁡ P ↾ 1 ..^ F -1 ∧ P 0 F ∩ P 1 ..^ F = ∅
20 17 18 19 3imtr4i ⊢ F SPaths ⁡ G P → F Paths ⁡ G P