Metamath Proof Explorer


Theorem efgredlema

Description: The reduced word that forms the base of the sequence in efgsval is uniquely determined, given the ending representation. (Contributed by Mario Carneiro, 1-Oct-2015)

Ref Expression
Hypotheses efgval.w W = I Word I × 2 𝑜
efgval.r ˙ = ~ FG I
efgval2.m M = y I , z 2 𝑜 y 1 𝑜 z
efgval2.t T = v W n 0 v , w I × 2 𝑜 v splice n n ⟨“ w M w ”⟩
efgred.d D = W x W ran T x
efgred.s S = m t Word W | t 0 D k 1 ..^ t t k ran T t k 1 m m 1
efgredlem.1 φ a dom S b dom S S a < S A S a = S b a 0 = b 0
efgredlem.2 φ A dom S
efgredlem.3 φ B dom S
efgredlem.4 φ S A = S B
efgredlem.5 φ ¬ A 0 = B 0
Assertion efgredlema φ A 1 B 1

Proof

Step Hyp Ref Expression
1 efgval.w W = I Word I × 2 𝑜
2 efgval.r ˙ = ~ FG I
3 efgval2.m M = y I , z 2 𝑜 y 1 𝑜 z
4 efgval2.t T = v W n 0 v , w I × 2 𝑜 v splice n n ⟨“ w M w ”⟩
5 efgred.d D = W x W ran T x
6 efgred.s S = m t Word W | t 0 D k 1 ..^ t t k ran T t k 1 m m 1
7 efgredlem.1 φ a dom S b dom S S a < S A S a = S b a 0 = b 0
8 efgredlem.2 φ A dom S
9 efgredlem.3 φ B dom S
10 efgredlem.4 φ S A = S B
11 efgredlem.5 φ ¬ A 0 = B 0
12 1 2 3 4 5 6 efgsval B dom S S B = B B 1
13 9 12 syl φ S B = B B 1
14 1 2 3 4 5 6 efgsval A dom S S A = A A 1
15 8 14 syl φ S A = A A 1
16 10 15 eqtr3d φ S B = A A 1
17 13 16 eqtr3d φ B B 1 = A A 1
18 oveq1 A = 1 A 1 = 1 1
19 1m1e0 1 1 = 0
20 18 19 eqtrdi A = 1 A 1 = 0
21 20 fveq2d A = 1 A A 1 = A 0
22 17 21 sylan9eq φ A = 1 B B 1 = A 0
23 10 eleq1d φ S A D S B D
24 1 2 3 4 5 6 efgs1b A dom S S A D A = 1
25 8 24 syl φ S A D A = 1
26 1 2 3 4 5 6 efgs1b B dom S S B D B = 1
27 9 26 syl φ S B D B = 1
28 23 25 27 3bitr3d φ A = 1 B = 1
29 28 biimpa φ A = 1 B = 1
30 oveq1 B = 1 B 1 = 1 1
31 30 19 eqtrdi B = 1 B 1 = 0
32 31 fveq2d B = 1 B B 1 = B 0
33 29 32 syl φ A = 1 B B 1 = B 0
34 22 33 eqtr3d φ A = 1 A 0 = B 0
35 11 34 mtand φ ¬ A = 1
36 1 2 3 4 5 6 efgsdm A dom S A Word W A 0 D u 1 ..^ A A u ran T A u 1
37 36 simp1bi A dom S A Word W
38 eldifsn A Word W A Word W A
39 lennncl A Word W A A
40 38 39 sylbi A Word W A
41 8 37 40 3syl φ A
42 elnn1uz2 A A = 1 A 2
43 41 42 sylib φ A = 1 A 2
44 43 ord φ ¬ A = 1 A 2
45 35 44 mpd φ A 2
46 uz2m1nn A 2 A 1
47 45 46 syl φ A 1
48 35 28 mtbid φ ¬ B = 1
49 1 2 3 4 5 6 efgsdm B dom S B Word W B 0 D u 1 ..^ B B u ran T B u 1
50 49 simp1bi B dom S B Word W
51 eldifsn B Word W B Word W B
52 lennncl B Word W B B
53 51 52 sylbi B Word W B
54 9 50 53 3syl φ B
55 elnn1uz2 B B = 1 B 2
56 54 55 sylib φ B = 1 B 2
57 56 ord φ ¬ B = 1 B 2
58 48 57 mpd φ B 2
59 uz2m1nn B 2 B 1
60 58 59 syl φ B 1
61 47 60 jca φ A 1 B 1