Metamath Proof Explorer


Theorem frgpuptinv

Description: Any assignment of the generators to target elements can be extended (uniquely) to a homomorphism from a free monoid to an arbitrary other monoid. (Contributed by Mario Carneiro, 2-Oct-2015)

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
frgpuptinv.m M = y I , z 2 𝑜 y 1 𝑜 z
Assertion frgpuptinv φ A I × 2 𝑜 T M A = N T 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 frgpuptinv.m M = y I , z 2 𝑜 y 1 𝑜 z
8 elxp2 A I × 2 𝑜 a I b 2 𝑜 A = a b
9 7 efgmval a I b 2 𝑜 a M b = a 1 𝑜 b
10 9 adantl φ a I b 2 𝑜 a M b = a 1 𝑜 b
11 10 fveq2d φ a I b 2 𝑜 T a M b = T a 1 𝑜 b
12 df-ov a T 1 𝑜 b = T a 1 𝑜 b
13 11 12 eqtr4di φ a I b 2 𝑜 T a M b = a T 1 𝑜 b
14 elpri b 1 𝑜 b = b = 1 𝑜
15 df2o3 2 𝑜 = 1 𝑜
16 14 15 eleq2s b 2 𝑜 b = b = 1 𝑜
17 simpr φ a I a I
18 1oelpr 1 𝑜 1 𝑜
19 18 15 eleqtrri 1 𝑜 2 𝑜
20 1n0 1 𝑜
21 neeq1 z = 1 𝑜 z 1 𝑜
22 20 21 mpbiri z = 1 𝑜 z
23 ifnefalse z if z = F y N F y = N F y
24 22 23 syl z = 1 𝑜 if z = F y N F y = N F y
25 fveq2 y = a F y = F a
26 25 fveq2d y = a N F y = N F a
27 24 26 sylan9eqr y = a z = 1 𝑜 if z = F y N F y = N F a
28 fvex N F a V
29 27 3 28 ovmpoa a I 1 𝑜 2 𝑜 a T 1 𝑜 = N F a
30 17 19 29 sylancl φ a I a T 1 𝑜 = N F a
31 0ex V
32 31 prid1 1 𝑜
33 32 15 eleqtrri 2 𝑜
34 iftrue z = if z = F y N F y = F y
35 34 25 sylan9eqr y = a z = if z = F y N F y = F a
36 fvex F a V
37 35 3 36 ovmpoa a I 2 𝑜 a T = F a
38 17 33 37 sylancl φ a I a T = F a
39 38 fveq2d φ a I N a T = N F a
40 30 39 eqtr4d φ a I a T 1 𝑜 = N a T
41 difeq2 b = 1 𝑜 b = 1 𝑜
42 dif0 1 𝑜 = 1 𝑜
43 41 42 eqtrdi b = 1 𝑜 b = 1 𝑜
44 43 oveq2d b = a T 1 𝑜 b = a T 1 𝑜
45 oveq2 b = a T b = a T
46 45 fveq2d b = N a T b = N a T
47 44 46 eqeq12d b = a T 1 𝑜 b = N a T b a T 1 𝑜 = N a T
48 40 47 syl5ibrcom φ a I b = a T 1 𝑜 b = N a T b
49 40 fveq2d φ a I N a T 1 𝑜 = N N a T
50 6 ffvelcdmda φ a I F a B
51 38 50 eqeltrd φ a I a T B
52 1 2 grpinvinv H Grp a T B N N a T = a T
53 4 51 52 syl2an2r φ a I N N a T = a T
54 49 53 eqtr2d φ a I a T = N a T 1 𝑜
55 difeq2 b = 1 𝑜 1 𝑜 b = 1 𝑜 1 𝑜
56 difid 1 𝑜 1 𝑜 =
57 55 56 eqtrdi b = 1 𝑜 1 𝑜 b =
58 57 oveq2d b = 1 𝑜 a T 1 𝑜 b = a T
59 oveq2 b = 1 𝑜 a T b = a T 1 𝑜
60 59 fveq2d b = 1 𝑜 N a T b = N a T 1 𝑜
61 58 60 eqeq12d b = 1 𝑜 a T 1 𝑜 b = N a T b a T = N a T 1 𝑜
62 54 61 syl5ibrcom φ a I b = 1 𝑜 a T 1 𝑜 b = N a T b
63 48 62 jaod φ a I b = b = 1 𝑜 a T 1 𝑜 b = N a T b
64 16 63 syl5 φ a I b 2 𝑜 a T 1 𝑜 b = N a T b
65 64 impr φ a I b 2 𝑜 a T 1 𝑜 b = N a T b
66 13 65 eqtrd φ a I b 2 𝑜 T a M b = N a T b
67 fveq2 A = a b M A = M a b
68 df-ov a M b = M a b
69 67 68 eqtr4di A = a b M A = a M b
70 69 fveq2d A = a b T M A = T a M b
71 fveq2 A = a b T A = T a b
72 df-ov a T b = T a b
73 71 72 eqtr4di A = a b T A = a T b
74 73 fveq2d A = a b N T A = N a T b
75 70 74 eqeq12d A = a b T M A = N T A T a M b = N a T b
76 66 75 syl5ibrcom φ a I b 2 𝑜 A = a b T M A = N T A
77 76 rexlimdvva φ a I b 2 𝑜 A = a b T M A = N T A
78 8 77 biimtrid φ A I × 2 𝑜 T M A = N T A
79 78 imp φ A I × 2 𝑜 T M A = N T A