Metamath Proof Explorer


Theorem frgpup2

Description: The evaluation map has the intended behavior on the generators. (Contributed by Mario Carneiro, 2-Oct-2015) (Revised by Mario Carneiro, 28-Feb-2016)

Ref Expression
Hypotheses frgpup.b B = Base H
frgpup.n N = inv g H
frgpup.t T = y I , z 2 𝑜 if z = F y N F y
frgpup.h φ H Grp
frgpup.i φ I V
frgpup.a φ F : I B
frgpup.w W = I Word I × 2 𝑜
frgpup.r ˙ = ~ FG I
frgpup.g G = freeGrp I
frgpup.x X = Base G
frgpup.e E = ran g W g ˙ H T g
frgpup.u U = var FGrp I
frgpup.y φ A I
Assertion frgpup2 φ E U A = F A

Proof

Step Hyp Ref Expression
1 frgpup.b B = Base H
2 frgpup.n N = inv g H
3 frgpup.t T = y I , z 2 𝑜 if z = F y N F y
4 frgpup.h φ H Grp
5 frgpup.i φ I V
6 frgpup.a φ F : I B
7 frgpup.w W = I Word I × 2 𝑜
8 frgpup.r ˙ = ~ FG I
9 frgpup.g G = freeGrp I
10 frgpup.x X = Base G
11 frgpup.e E = ran g W g ˙ H T g
12 frgpup.u U = var FGrp I
13 frgpup.y φ A I
14 8 12 vrgpval I V A I U A = ⟨“ A ”⟩ ˙
15 5 13 14 syl2anc φ U A = ⟨“ A ”⟩ ˙
16 15 fveq2d φ E U A = E ⟨“ A ”⟩ ˙
17 0ex V
18 17 prid1 1 𝑜
19 df2o3 2 𝑜 = 1 𝑜
20 18 19 eleqtrri 2 𝑜
21 opelxpi A I 2 𝑜 A I × 2 𝑜
22 13 20 21 sylancl φ A I × 2 𝑜
23 22 s1cld φ ⟨“ A ”⟩ Word I × 2 𝑜
24 2on 2 𝑜 On
25 xpexg I V 2 𝑜 On I × 2 𝑜 V
26 5 24 25 sylancl φ I × 2 𝑜 V
27 wrdexg I × 2 𝑜 V Word I × 2 𝑜 V
28 fvi Word I × 2 𝑜 V I Word I × 2 𝑜 = Word I × 2 𝑜
29 26 27 28 3syl φ I Word I × 2 𝑜 = Word I × 2 𝑜
30 7 29 syl5eq φ W = Word I × 2 𝑜
31 23 30 eleqtrrd φ ⟨“ A ”⟩ W
32 1 2 3 4 5 6 7 8 9 10 11 frgpupval φ ⟨“ A ”⟩ W E ⟨“ A ”⟩ ˙ = H T ⟨“ A ”⟩
33 31 32 mpdan φ E ⟨“ A ”⟩ ˙ = H T ⟨“ A ”⟩
34 1 2 3 4 5 6 frgpuptf φ T : I × 2 𝑜 B
35 s1co A I × 2 𝑜 T : I × 2 𝑜 B T ⟨“ A ”⟩ = ⟨“ T A ”⟩
36 22 34 35 syl2anc φ T ⟨“ A ”⟩ = ⟨“ T A ”⟩
37 df-ov A T = T A
38 iftrue z = if z = F y N F y = F y
39 fveq2 y = A F y = F A
40 38 39 sylan9eqr y = A z = if z = F y N F y = F A
41 fvex F A V
42 40 3 41 ovmpoa A I 2 𝑜 A T = F A
43 13 20 42 sylancl φ A T = F A
44 37 43 syl5eqr φ T A = F A
45 44 s1eqd φ ⟨“ T A ”⟩ = ⟨“ F A ”⟩
46 36 45 eqtrd φ T ⟨“ A ”⟩ = ⟨“ F A ”⟩
47 46 oveq2d φ H T ⟨“ A ”⟩ = H ⟨“ F A ”⟩
48 6 13 ffvelrnd φ F A B
49 1 gsumws1 F A B H ⟨“ F A ”⟩ = F A
50 48 49 syl φ H ⟨“ F A ”⟩ = F A
51 47 50 eqtrd φ H T ⟨“ A ”⟩ = F A
52 16 33 51 3eqtrd φ E U A = F A