Metamath Proof Explorer


Theorem nminv

Description: The norm of a negated element is the same as the norm of the original element. (Contributed by Mario Carneiro, 4-Oct-2015)

Ref Expression
Hypotheses nmf.x 𝑋 = ( Base ‘ 𝐺 )
nmf.n 𝑁 = ( norm ‘ 𝐺 )
nminv.i 𝐼 = ( invg𝐺 )
Assertion nminv ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ) → ( 𝑁 ‘ ( 𝐼𝐴 ) ) = ( 𝑁𝐴 ) )

Proof

Step Hyp Ref Expression
1 nmf.x 𝑋 = ( Base ‘ 𝐺 )
2 nmf.n 𝑁 = ( norm ‘ 𝐺 )
3 nminv.i 𝐼 = ( invg𝐺 )
4 ngpgrp ( 𝐺 ∈ NrmGrp → 𝐺 ∈ Grp )
5 4 adantr ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ) → 𝐺 ∈ Grp )
6 eqid ( 0g𝐺 ) = ( 0g𝐺 )
7 1 6 grpidcl ( 𝐺 ∈ Grp → ( 0g𝐺 ) ∈ 𝑋 )
8 5 7 syl ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ) → ( 0g𝐺 ) ∈ 𝑋 )
9 eqid ( -g𝐺 ) = ( -g𝐺 )
10 eqid ( dist ‘ 𝐺 ) = ( dist ‘ 𝐺 )
11 2 1 9 10 ngpdsr ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ∧ ( 0g𝐺 ) ∈ 𝑋 ) → ( 𝐴 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) = ( 𝑁 ‘ ( ( 0g𝐺 ) ( -g𝐺 ) 𝐴 ) ) )
12 8 11 mpd3an3 ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ) → ( 𝐴 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) = ( 𝑁 ‘ ( ( 0g𝐺 ) ( -g𝐺 ) 𝐴 ) ) )
13 2 1 6 10 nmval ( 𝐴𝑋 → ( 𝑁𝐴 ) = ( 𝐴 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) )
14 13 adantl ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ) → ( 𝑁𝐴 ) = ( 𝐴 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) )
15 1 9 3 6 grpinvval2 ( ( 𝐺 ∈ Grp ∧ 𝐴𝑋 ) → ( 𝐼𝐴 ) = ( ( 0g𝐺 ) ( -g𝐺 ) 𝐴 ) )
16 4 15 sylan ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ) → ( 𝐼𝐴 ) = ( ( 0g𝐺 ) ( -g𝐺 ) 𝐴 ) )
17 16 fveq2d ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ) → ( 𝑁 ‘ ( 𝐼𝐴 ) ) = ( 𝑁 ‘ ( ( 0g𝐺 ) ( -g𝐺 ) 𝐴 ) ) )
18 12 14 17 3eqtr4rd ( ( 𝐺 ∈ NrmGrp ∧ 𝐴𝑋 ) → ( 𝑁 ‘ ( 𝐼𝐴 ) ) = ( 𝑁𝐴 ) )