Metamath Proof Explorer


Theorem lfgrwlkprop

Description: Two adjacent vertices in a walk are different in a loop-free graph. (Contributed by AV, 28-Jan-2021)

Ref Expression
Hypothesis lfgrwlkprop.i I = iEdg G
Assertion lfgrwlkprop F Walks G P I : dom I x 𝒫 V | 2 x k 0 ..^ F P k P k + 1

Proof

Step Hyp Ref Expression
1 lfgrwlkprop.i I = iEdg G
2 wlkv F Walks G P G V F V P V
3 eqid Vtx G = Vtx G
4 3 1 iswlk G V F V P V F Walks G P F Word dom I P : 0 F Vtx G k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k
5 2 4 syl F Walks G P F Walks G P F Word dom I P : 0 F Vtx G k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k
6 ifptru P k = P k + 1 if- P k = P k + 1 I F k = P k P k P k + 1 I F k I F k = P k
7 6 adantr P k = P k + 1 F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k I F k = P k
8 simplr F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F I : dom I x 𝒫 V | 2 x
9 wrdsymbcl F Word dom I k 0 ..^ F F k dom I
10 9 ad4ant14 F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F F k dom I
11 8 10 ffvelrnd F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F I F k x 𝒫 V | 2 x
12 fveq2 x = I F k x = I F k
13 12 breq2d x = I F k 2 x 2 I F k
14 13 elrab I F k x 𝒫 V | 2 x I F k 𝒫 V 2 I F k
15 fveq2 I F k = P k I F k = P k
16 15 breq2d I F k = P k 2 I F k 2 P k
17 fvex P k V
18 hashsng P k V P k = 1
19 17 18 ax-mp P k = 1
20 19 breq2i 2 P k 2 1
21 1lt2 1 < 2
22 1re 1
23 2re 2
24 22 23 ltnlei 1 < 2 ¬ 2 1
25 pm2.21 ¬ 2 1 2 1 P k P k + 1
26 24 25 sylbi 1 < 2 2 1 P k P k + 1
27 21 26 ax-mp 2 1 P k P k + 1
28 20 27 sylbi 2 P k P k P k + 1
29 16 28 syl6bi I F k = P k 2 I F k P k P k + 1
30 29 com12 2 I F k I F k = P k P k P k + 1
31 30 adantl I F k 𝒫 V 2 I F k I F k = P k P k P k + 1
32 31 a1i F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F I F k 𝒫 V 2 I F k I F k = P k P k P k + 1
33 14 32 syl5bi F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F I F k x 𝒫 V | 2 x I F k = P k P k P k + 1
34 11 33 mpd F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F I F k = P k P k P k + 1
35 34 adantl P k = P k + 1 F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F I F k = P k P k P k + 1
36 7 35 sylbid P k = P k + 1 F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k P k P k + 1
37 36 ex P k = P k + 1 F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k P k P k + 1
38 neqne ¬ P k = P k + 1 P k P k + 1
39 38 2a1d ¬ P k = P k + 1 F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k P k P k + 1
40 37 39 pm2.61i F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k P k P k + 1
41 40 ralimdva F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k k 0 ..^ F P k P k + 1
42 41 ex F Word dom I P : 0 F Vtx G I : dom I x 𝒫 V | 2 x k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k k 0 ..^ F P k P k + 1
43 42 com23 F Word dom I P : 0 F Vtx G k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k I : dom I x 𝒫 V | 2 x k 0 ..^ F P k P k + 1
44 43 3impia F Word dom I P : 0 F Vtx G k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k I : dom I x 𝒫 V | 2 x k 0 ..^ F P k P k + 1
45 5 44 syl6bi F Walks G P F Walks G P I : dom I x 𝒫 V | 2 x k 0 ..^ F P k P k + 1
46 45 pm2.43i F Walks G P I : dom I x 𝒫 V | 2 x k 0 ..^ F P k P k + 1
47 46 imp F Walks G P I : dom I x 𝒫 V | 2 x k 0 ..^ F P k P k + 1