Metamath Proof Explorer


Theorem efgred2

Description: Two extension sequences have related endpoints iff they have the same base. (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
Assertion efgred2 A dom S B dom S S A ˙ S B A 0 = B 0

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 1 2 3 4 5 6 efgsfo S : dom S onto W
8 fof S : dom S onto W S : dom S W
9 7 8 ax-mp S : dom S W
10 9 ffvelrni B dom S S B W
11 10 ad2antlr A dom S B dom S S A ˙ S B S B W
12 1 2 3 4 5 6 efgredeu S B W ∃! d D d ˙ S B
13 reurmo ∃! d D d ˙ S B * d D d ˙ S B
14 11 12 13 3syl A dom S B dom S S A ˙ S B * d D d ˙ S B
15 1 2 3 4 5 6 efgsdm A dom S A Word W A 0 D i 1 ..^ A A i ran T A i 1
16 15 simp2bi A dom S A 0 D
17 16 ad2antrr A dom S B dom S S A ˙ S B A 0 D
18 1 2 efger ˙ Er W
19 18 a1i A dom S B dom S S A ˙ S B ˙ Er W
20 1 2 3 4 5 6 efgsrel A dom S A 0 ˙ S A
21 20 ad2antrr A dom S B dom S S A ˙ S B A 0 ˙ S A
22 simpr A dom S B dom S S A ˙ S B S A ˙ S B
23 19 21 22 ertrd A dom S B dom S S A ˙ S B A 0 ˙ S B
24 1 2 3 4 5 6 efgsdm B dom S B Word W B 0 D i 1 ..^ B B i ran T B i 1
25 24 simp2bi B dom S B 0 D
26 25 ad2antlr A dom S B dom S S A ˙ S B B 0 D
27 1 2 3 4 5 6 efgsrel B dom S B 0 ˙ S B
28 27 ad2antlr A dom S B dom S S A ˙ S B B 0 ˙ S B
29 breq1 d = A 0 d ˙ S B A 0 ˙ S B
30 breq1 d = B 0 d ˙ S B B 0 ˙ S B
31 29 30 rmoi * d D d ˙ S B A 0 D A 0 ˙ S B B 0 D B 0 ˙ S B A 0 = B 0
32 14 17 23 26 28 31 syl122anc A dom S B dom S S A ˙ S B A 0 = B 0
33 18 a1i A dom S B dom S A 0 = B 0 ˙ Er W
34 20 ad2antrr A dom S B dom S A 0 = B 0 A 0 ˙ S A
35 simpr A dom S B dom S A 0 = B 0 A 0 = B 0
36 27 ad2antlr A dom S B dom S A 0 = B 0 B 0 ˙ S B
37 35 36 eqbrtrd A dom S B dom S A 0 = B 0 A 0 ˙ S B
38 33 34 37 ertr3d A dom S B dom S A 0 = B 0 S A ˙ S B
39 32 38 impbida A dom S B dom S S A ˙ S B A 0 = B 0