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