Metamath Proof Explorer


Theorem 1wlkdlem4

Description: Lemma 4 for 1wlkd . (Contributed by AV, 22-Jan-2021)

Ref Expression
Hypotheses 1wlkd.p P = ⟨“ XY ”⟩
1wlkd.f F = ⟨“ J ”⟩
1wlkd.x φ X V
1wlkd.y φ Y V
1wlkd.l φ X = Y I J = X
1wlkd.j φ X Y X Y I J
Assertion 1wlkdlem4 φ k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k

Proof

Step Hyp Ref Expression
1 1wlkd.p P = ⟨“ XY ”⟩
2 1wlkd.f F = ⟨“ J ”⟩
3 1wlkd.x φ X V
4 1wlkd.y φ Y V
5 1wlkd.l φ X = Y I J = X
6 1wlkd.j φ X Y X Y I J
7 2 fveq1i F 0 = ⟨“ J ”⟩ 0
8 1 2 3 4 5 6 1wlkdlem2 φ X I J
9 8 elfvexd φ J V
10 s1fv J V ⟨“ J ”⟩ 0 = J
11 9 10 syl φ ⟨“ J ”⟩ 0 = J
12 7 11 eqtrid φ F 0 = J
13 12 fveq2d φ I F 0 = I J
14 13 adantr φ X = Y I F 0 = I J
15 14 5 eqtrd φ X = Y I F 0 = X
16 df-ne X Y ¬ X = Y
17 16 6 sylan2br φ ¬ X = Y X Y I J
18 13 adantr φ ¬ X = Y I F 0 = I J
19 17 18 sseqtrrd φ ¬ X = Y X Y I F 0
20 15 19 ifpimpda φ if- X = Y I F 0 = X X Y I F 0
21 1 fveq1i P 0 = ⟨“ XY ”⟩ 0
22 s2fv0 X V ⟨“ XY ”⟩ 0 = X
23 3 22 syl φ ⟨“ XY ”⟩ 0 = X
24 21 23 eqtrid φ P 0 = X
25 1 fveq1i P 1 = ⟨“ XY ”⟩ 1
26 s2fv1 Y V ⟨“ XY ”⟩ 1 = Y
27 4 26 syl φ ⟨“ XY ”⟩ 1 = Y
28 25 27 eqtrid φ P 1 = Y
29 eqeq12 P 0 = X P 1 = Y P 0 = P 1 X = Y
30 sneq P 0 = X P 0 = X
31 30 adantr P 0 = X P 1 = Y P 0 = X
32 31 eqeq2d P 0 = X P 1 = Y I F 0 = P 0 I F 0 = X
33 preq12 P 0 = X P 1 = Y P 0 P 1 = X Y
34 33 sseq1d P 0 = X P 1 = Y P 0 P 1 I F 0 X Y I F 0
35 29 32 34 ifpbi123d P 0 = X P 1 = Y if- P 0 = P 1 I F 0 = P 0 P 0 P 1 I F 0 if- X = Y I F 0 = X X Y I F 0
36 24 28 35 syl2anc φ if- P 0 = P 1 I F 0 = P 0 P 0 P 1 I F 0 if- X = Y I F 0 = X X Y I F 0
37 20 36 mpbird φ if- P 0 = P 1 I F 0 = P 0 P 0 P 1 I F 0
38 c0ex 0 V
39 oveq1 k = 0 k + 1 = 0 + 1
40 0p1e1 0 + 1 = 1
41 39 40 eqtrdi k = 0 k + 1 = 1
42 wkslem2 k = 0 k + 1 = 1 if- P k = P k + 1 I F k = P k P k P k + 1 I F k if- P 0 = P 1 I F 0 = P 0 P 0 P 1 I F 0
43 41 42 mpdan k = 0 if- P k = P k + 1 I F k = P k P k P k + 1 I F k if- P 0 = P 1 I F 0 = P 0 P 0 P 1 I F 0
44 38 43 ralsn k 0 if- P k = P k + 1 I F k = P k P k P k + 1 I F k if- P 0 = P 1 I F 0 = P 0 P 0 P 1 I F 0
45 37 44 sylibr φ k 0 if- P k = P k + 1 I F k = P k P k P k + 1 I F k
46 2 fveq2i F = ⟨“ J ”⟩
47 s1len ⟨“ J ”⟩ = 1
48 46 47 eqtri F = 1
49 48 oveq2i 0 ..^ F = 0 ..^ 1
50 fzo01 0 ..^ 1 = 0
51 49 50 eqtri 0 ..^ F = 0
52 51 a1i φ 0 ..^ F = 0
53 52 raleqdv φ k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k k 0 if- P k = P k + 1 I F k = P k P k P k + 1 I F k
54 45 53 mpbird φ k 0 ..^ F if- P k = P k + 1 I F k = P k P k P k + 1 I F k