Metamath Proof Explorer


Theorem crctcshlem4

Description: Lemma for crctcsh . (Contributed by AV, 10-Mar-2021)

Ref Expression
Hypotheses crctcsh.v V = Vtx G
crctcsh.i I = iEdg G
crctcsh.d φ F Circuits G P
crctcsh.n N = F
crctcsh.s φ S 0 ..^ N
crctcsh.h H = F cyclShift S
crctcsh.q Q = x 0 N if x N S P x + S P x + S - N
Assertion crctcshlem4 φ S = 0 H = F Q = P

Proof

Step Hyp Ref Expression
1 crctcsh.v V = Vtx G
2 crctcsh.i I = iEdg G
3 crctcsh.d φ F Circuits G P
4 crctcsh.n N = F
5 crctcsh.s φ S 0 ..^ N
6 crctcsh.h H = F cyclShift S
7 crctcsh.q Q = x 0 N if x N S P x + S P x + S - N
8 oveq2 S = 0 F cyclShift S = F cyclShift 0
9 crctiswlk F Circuits G P F Walks G P
10 2 wlkf F Walks G P F Word dom I
11 3 9 10 3syl φ F Word dom I
12 cshw0 F Word dom I F cyclShift 0 = F
13 11 12 syl φ F cyclShift 0 = F
14 8 13 sylan9eqr φ S = 0 F cyclShift S = F
15 6 14 eqtrid φ S = 0 H = F
16 oveq2 S = 0 N S = N 0
17 1 2 3 4 crctcshlem1 φ N 0
18 17 nn0cnd φ N
19 18 subid1d φ N 0 = N
20 16 19 sylan9eqr φ S = 0 N S = N
21 20 breq2d φ S = 0 x N S x N
22 21 adantr φ S = 0 x 0 N x N S x N
23 oveq2 S = 0 x + S = x + 0
24 23 adantl φ S = 0 x + S = x + 0
25 elfzelz x 0 N x
26 25 zcnd x 0 N x
27 26 addid1d x 0 N x + 0 = x
28 24 27 sylan9eq φ S = 0 x 0 N x + S = x
29 28 fveq2d φ S = 0 x 0 N P x + S = P x
30 28 fvoveq1d φ S = 0 x 0 N P x + S - N = P x N
31 22 29 30 ifbieq12d φ S = 0 x 0 N if x N S P x + S P x + S - N = if x N P x P x N
32 31 mpteq2dva φ S = 0 x 0 N if x N S P x + S P x + S - N = x 0 N if x N P x P x N
33 elfzle2 x 0 N x N
34 33 adantl φ x 0 N x N
35 34 iftrued φ x 0 N if x N P x P x N = P x
36 35 mpteq2dva φ x 0 N if x N P x P x N = x 0 N P x
37 1 wlkp F Walks G P P : 0 F V
38 3 9 37 3syl φ P : 0 F V
39 ffn P : 0 F V P Fn 0 F
40 4 eqcomi F = N
41 40 oveq2i 0 F = 0 N
42 41 fneq2i P Fn 0 F P Fn 0 N
43 39 42 sylib P : 0 F V P Fn 0 N
44 43 adantl φ P : 0 F V P Fn 0 N
45 dffn5 P Fn 0 N P = x 0 N P x
46 44 45 sylib φ P : 0 F V P = x 0 N P x
47 46 eqcomd φ P : 0 F V x 0 N P x = P
48 38 47 mpdan φ x 0 N P x = P
49 36 48 eqtrd φ x 0 N if x N P x P x N = P
50 49 adantr φ S = 0 x 0 N if x N P x P x N = P
51 32 50 eqtrd φ S = 0 x 0 N if x N S P x + S P x + S - N = P
52 7 51 eqtrid φ S = 0 Q = P
53 15 52 jca φ S = 0 H = F Q = P