Metamath Proof Explorer


Theorem trlsegvdeglem2

Description: Lemma for trlsegvdeg . (Contributed by AV, 20-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
Assertion trlsegvdeglem2 ⊢ φ → Fun ⁡ iEdg ⁡ X

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 3 funresd ⊢ φ → Fun ⁡ I ↾ F 0 ..^ N
14 10 funeqd ⊢ φ → Fun ⁡ iEdg ⁡ X ↔ Fun ⁡ I ↾ F 0 ..^ N
15 13 14 mpbird ⊢ φ → Fun ⁡ iEdg ⁡ X