Metamath Proof Explorer


Theorem nvm1

Description: The norm of the negative of a vector. (Contributed by NM, 28-Nov-2006) (New usage is discouraged.)

Ref Expression
Hypotheses nvs.1 𝑋 = ( BaseSet ‘ 𝑈 )
nvs.4 𝑆 = ( ·𝑠OLD𝑈 )
nvs.6 𝑁 = ( normCV𝑈 )
Assertion nvm1 ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( 𝑁 ‘ ( - 1 𝑆 𝐴 ) ) = ( 𝑁𝐴 ) )

Proof

Step Hyp Ref Expression
1 nvs.1 𝑋 = ( BaseSet ‘ 𝑈 )
2 nvs.4 𝑆 = ( ·𝑠OLD𝑈 )
3 nvs.6 𝑁 = ( normCV𝑈 )
4 neg1cn - 1 ∈ ℂ
5 1 2 3 nvs ( ( 𝑈 ∈ NrmCVec ∧ - 1 ∈ ℂ ∧ 𝐴𝑋 ) → ( 𝑁 ‘ ( - 1 𝑆 𝐴 ) ) = ( ( abs ‘ - 1 ) · ( 𝑁𝐴 ) ) )
6 4 5 mp3an2 ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( 𝑁 ‘ ( - 1 𝑆 𝐴 ) ) = ( ( abs ‘ - 1 ) · ( 𝑁𝐴 ) ) )
7 ax-1cn 1 ∈ ℂ
8 7 absnegi ( abs ‘ - 1 ) = ( abs ‘ 1 )
9 abs1 ( abs ‘ 1 ) = 1
10 8 9 eqtri ( abs ‘ - 1 ) = 1
11 10 oveq1i ( ( abs ‘ - 1 ) · ( 𝑁𝐴 ) ) = ( 1 · ( 𝑁𝐴 ) )
12 1 3 nvcl ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( 𝑁𝐴 ) ∈ ℝ )
13 12 recnd ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( 𝑁𝐴 ) ∈ ℂ )
14 13 mulid2d ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( 1 · ( 𝑁𝐴 ) ) = ( 𝑁𝐴 ) )
15 11 14 syl5eq ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( ( abs ‘ - 1 ) · ( 𝑁𝐴 ) ) = ( 𝑁𝐴 ) )
16 6 15 eqtrd ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( 𝑁 ‘ ( - 1 𝑆 𝐴 ) ) = ( 𝑁𝐴 ) )