Metamath Proof Explorer


Theorem normsq

Description: The square of a norm. (Contributed by NM, 12-May-2005) (New usage is discouraged.)

Ref Expression
Assertion normsq ( 𝐴 ∈ ℋ → ( ( norm𝐴 ) ↑ 2 ) = ( 𝐴 ·ih 𝐴 ) )

Proof

Step Hyp Ref Expression
1 fveq2 ( 𝐴 = if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) → ( norm𝐴 ) = ( norm ‘ if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ) )
2 1 oveq1d ( 𝐴 = if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) → ( ( norm𝐴 ) ↑ 2 ) = ( ( norm ‘ if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ) ↑ 2 ) )
3 id ( 𝐴 = if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) → 𝐴 = if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) )
4 3 3 oveq12d ( 𝐴 = if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) → ( 𝐴 ·ih 𝐴 ) = ( if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ·ih if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ) )
5 2 4 eqeq12d ( 𝐴 = if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) → ( ( ( norm𝐴 ) ↑ 2 ) = ( 𝐴 ·ih 𝐴 ) ↔ ( ( norm ‘ if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ) ↑ 2 ) = ( if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ·ih if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ) ) )
6 ifhvhv0 if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ∈ ℋ
7 6 normsqi ( ( norm ‘ if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ) ↑ 2 ) = ( if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) ·ih if ( 𝐴 ∈ ℋ , 𝐴 , 0 ) )
8 5 7 dedth ( 𝐴 ∈ ℋ → ( ( norm𝐴 ) ↑ 2 ) = ( 𝐴 ·ih 𝐴 ) )