Metamath Proof Explorer


Theorem frgpinv

Description: The inverse of an element of the free group. (Contributed by Mario Carneiro, 2-Oct-2015)

Ref Expression
Hypotheses frgpadd.w W=IWordI×2𝑜
frgpadd.g G=freeGrpI
frgpadd.r ˙=~FGI
frgpinv.n N=invgG
frgpinv.m M=yI,z2𝑜y1𝑜z
Assertion frgpinv AWNA˙=MreverseA˙

Proof

Step Hyp Ref Expression
1 frgpadd.w W=IWordI×2𝑜
2 frgpadd.g G=freeGrpI
3 frgpadd.r ˙=~FGI
4 frgpinv.n N=invgG
5 frgpinv.m M=yI,z2𝑜y1𝑜z
6 fviss IWordI×2𝑜WordI×2𝑜
7 1 6 eqsstri WWordI×2𝑜
8 7 sseli AWAWordI×2𝑜
9 revcl AWordI×2𝑜reverseAWordI×2𝑜
10 8 9 syl AWreverseAWordI×2𝑜
11 5 efgmf M:I×2𝑜I×2𝑜
12 wrdco reverseAWordI×2𝑜M:I×2𝑜I×2𝑜MreverseAWordI×2𝑜
13 10 11 12 sylancl AWMreverseAWordI×2𝑜
14 1 efgrcl AWIVW=WordI×2𝑜
15 14 simprd AWW=WordI×2𝑜
16 13 15 eleqtrrd AWMreverseAW
17 eqid +G=+G
18 1 2 3 17 frgpadd AWMreverseAWA˙+GMreverseA˙=A++MreverseA˙
19 16 18 mpdan AWA˙+GMreverseA˙=A++MreverseA˙
20 1 3 efger ˙ErW
21 20 a1i AW˙ErW
22 eqid vWn0v,wI×2𝑜vsplicenn⟨“wMw”⟩=vWn0v,wI×2𝑜vsplicenn⟨“wMw”⟩
23 1 3 5 22 efginvrel2 AWA++MreverseA˙
24 21 23 erthi AWA++MreverseA˙=˙
25 2 3 frgp0 IVGGrp˙=0G
26 25 adantr IVW=WordI×2𝑜GGrp˙=0G
27 14 26 syl AWGGrp˙=0G
28 27 simprd AW˙=0G
29 19 24 28 3eqtrd AWA˙+GMreverseA˙=0G
30 27 simpld AWGGrp
31 eqid BaseG=BaseG
32 2 3 1 31 frgpeccl AWA˙BaseG
33 2 3 1 31 frgpeccl MreverseAWMreverseA˙BaseG
34 16 33 syl AWMreverseA˙BaseG
35 eqid 0G=0G
36 31 17 35 4 grpinvid1 GGrpA˙BaseGMreverseA˙BaseGNA˙=MreverseA˙A˙+GMreverseA˙=0G
37 30 32 34 36 syl3anc AWNA˙=MreverseA˙A˙+GMreverseA˙=0G
38 29 37 mpbird AWNA˙=MreverseA˙