Step |
Hyp |
Ref |
Expression |
1 |
|
ngprcan.x |
⊢ 𝑋 = ( Base ‘ 𝐺 ) |
2 |
|
ngprcan.p |
⊢ + = ( +g ‘ 𝐺 ) |
3 |
|
ngprcan.d |
⊢ 𝐷 = ( dist ‘ 𝐺 ) |
4 |
|
ngpgrp |
⊢ ( 𝐺 ∈ NrmGrp → 𝐺 ∈ Grp ) |
5 |
|
eqid |
⊢ ( -g ‘ 𝐺 ) = ( -g ‘ 𝐺 ) |
6 |
1 2 5
|
grppnpcan2 |
⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( 𝐴 + 𝐶 ) ( -g ‘ 𝐺 ) ( 𝐵 + 𝐶 ) ) = ( 𝐴 ( -g ‘ 𝐺 ) 𝐵 ) ) |
7 |
4 6
|
sylan |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( 𝐴 + 𝐶 ) ( -g ‘ 𝐺 ) ( 𝐵 + 𝐶 ) ) = ( 𝐴 ( -g ‘ 𝐺 ) 𝐵 ) ) |
8 |
7
|
fveq2d |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( norm ‘ 𝐺 ) ‘ ( ( 𝐴 + 𝐶 ) ( -g ‘ 𝐺 ) ( 𝐵 + 𝐶 ) ) ) = ( ( norm ‘ 𝐺 ) ‘ ( 𝐴 ( -g ‘ 𝐺 ) 𝐵 ) ) ) |
9 |
|
simpl |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → 𝐺 ∈ NrmGrp ) |
10 |
4
|
adantr |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → 𝐺 ∈ Grp ) |
11 |
|
simpr1 |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → 𝐴 ∈ 𝑋 ) |
12 |
|
simpr3 |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → 𝐶 ∈ 𝑋 ) |
13 |
1 2
|
grpcl |
⊢ ( ( 𝐺 ∈ Grp ∧ 𝐴 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) → ( 𝐴 + 𝐶 ) ∈ 𝑋 ) |
14 |
10 11 12 13
|
syl3anc |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( 𝐴 + 𝐶 ) ∈ 𝑋 ) |
15 |
|
simpr2 |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → 𝐵 ∈ 𝑋 ) |
16 |
1 2
|
grpcl |
⊢ ( ( 𝐺 ∈ Grp ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) → ( 𝐵 + 𝐶 ) ∈ 𝑋 ) |
17 |
10 15 12 16
|
syl3anc |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( 𝐵 + 𝐶 ) ∈ 𝑋 ) |
18 |
|
eqid |
⊢ ( norm ‘ 𝐺 ) = ( norm ‘ 𝐺 ) |
19 |
18 1 5 3
|
ngpds |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 + 𝐶 ) ∈ 𝑋 ∧ ( 𝐵 + 𝐶 ) ∈ 𝑋 ) → ( ( 𝐴 + 𝐶 ) 𝐷 ( 𝐵 + 𝐶 ) ) = ( ( norm ‘ 𝐺 ) ‘ ( ( 𝐴 + 𝐶 ) ( -g ‘ 𝐺 ) ( 𝐵 + 𝐶 ) ) ) ) |
20 |
9 14 17 19
|
syl3anc |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( 𝐴 + 𝐶 ) 𝐷 ( 𝐵 + 𝐶 ) ) = ( ( norm ‘ 𝐺 ) ‘ ( ( 𝐴 + 𝐶 ) ( -g ‘ 𝐺 ) ( 𝐵 + 𝐶 ) ) ) ) |
21 |
18 1 5 3
|
ngpds |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝐷 𝐵 ) = ( ( norm ‘ 𝐺 ) ‘ ( 𝐴 ( -g ‘ 𝐺 ) 𝐵 ) ) ) |
22 |
9 11 15 21
|
syl3anc |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( 𝐴 𝐷 𝐵 ) = ( ( norm ‘ 𝐺 ) ‘ ( 𝐴 ( -g ‘ 𝐺 ) 𝐵 ) ) ) |
23 |
8 20 22
|
3eqtr4d |
⊢ ( ( 𝐺 ∈ NrmGrp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( 𝐴 + 𝐶 ) 𝐷 ( 𝐵 + 𝐶 ) ) = ( 𝐴 𝐷 𝐵 ) ) |