Metamath Proof Explorer


Theorem eupth2lem3lem5

Description: Lemma for eupth2 . (Contributed by AV, 25-Feb-2021)

Ref Expression
Hypotheses trlsegvdeg.v ⊢ V = Vtx ⁡ G
trlsegvdeg.i ⊢ I = iEdg ⁡ G
trlsegvdeg.f ⊢ φ → Fun ⁡ I
trlsegvdeg.n ⊢ φ → N ∈ 0 ..^ F
trlsegvdeg.u ⊢ φ → U ∈ V
trlsegvdeg.w ⊢ φ → F Trails ⁡ G P
trlsegvdeg.vx ⊢ φ → Vtx ⁡ X = V
trlsegvdeg.vy ⊢ φ → Vtx ⁡ Y = V
trlsegvdeg.vz ⊢ φ → Vtx ⁡ Z = V
trlsegvdeg.ix ⊢ φ → iEdg ⁡ X = I ↾ F 0 ..^ N
trlsegvdeg.iy ⊢ φ → iEdg ⁡ Y = F ⁡ N I ⁡ F ⁡ N
trlsegvdeg.iz ⊢ φ → iEdg ⁡ Z = I ↾ F 0 … N
eupth2lem3.o ⊢ φ → x ∈ V | ¬ 2 ∥ VtxDeg ⁡ X ⁡ x = if P ⁡ 0 = P ⁡ N ∅ P ⁡ 0 P ⁡ N
eupth2lem3.e ⊢ φ → I ⁡ F ⁡ N = P ⁡ N P ⁡ N + 1
Assertion eupth2lem3lem5 ⊢ φ → I ⁡ F ⁡ N ∈ 𝒫 V

Proof

Step Hyp Ref Expression
1 trlsegvdeg.v ⊢ V = Vtx ⁡ G
2 trlsegvdeg.i ⊢ I = iEdg ⁡ G
3 trlsegvdeg.f ⊢ φ → Fun ⁡ I
4 trlsegvdeg.n ⊢ φ → N ∈ 0 ..^ F
5 trlsegvdeg.u ⊢ φ → U ∈ V
6 trlsegvdeg.w ⊢ φ → F Trails ⁡ G P
7 trlsegvdeg.vx ⊢ φ → Vtx ⁡ X = V
8 trlsegvdeg.vy ⊢ φ → Vtx ⁡ Y = V
9 trlsegvdeg.vz ⊢ φ → Vtx ⁡ Z = V
10 trlsegvdeg.ix ⊢ φ → iEdg ⁡ X = I ↾ F 0 ..^ N
11 trlsegvdeg.iy ⊢ φ → iEdg ⁡ Y = F ⁡ N I ⁡ F ⁡ N
12 trlsegvdeg.iz ⊢ φ → iEdg ⁡ Z = I ↾ F 0 … N
13 eupth2lem3.o ⊢ φ → x ∈ V | ¬ 2 ∥ VtxDeg ⁡ X ⁡ x = if P ⁡ 0 = P ⁡ N ∅ P ⁡ 0 P ⁡ N
14 eupth2lem3.e ⊢ φ → I ⁡ F ⁡ N = P ⁡ N P ⁡ N + 1
15 1 2 3 4 5 6 trlsegvdeglem1 ⊢ φ → P ⁡ N ∈ V ∧ P ⁡ N + 1 ∈ V
16 prelpwi ⊢ P ⁡ N ∈ V ∧ P ⁡ N + 1 ∈ V → P ⁡ N P ⁡ N + 1 ∈ 𝒫 V
17 15 16 syl ⊢ φ → P ⁡ N P ⁡ N + 1 ∈ 𝒫 V
18 14 17 eqeltrd ⊢ φ → I ⁡ F ⁡ N ∈ 𝒫 V