Metamath Proof Explorer


Theorem nvz

Description: The norm of a vector is zero iff the vector is zero. First part of Problem 2 of Kreyszig p. 64. (Contributed by NM, 24-Nov-2006) (New usage is discouraged.)

Ref Expression
Hypotheses nvz.1 𝑋 = ( BaseSet ‘ 𝑈 )
nvz.5 𝑍 = ( 0vec𝑈 )
nvz.6 𝑁 = ( normCV𝑈 )
Assertion nvz ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( ( 𝑁𝐴 ) = 0 ↔ 𝐴 = 𝑍 ) )

Proof

Step Hyp Ref Expression
1 nvz.1 𝑋 = ( BaseSet ‘ 𝑈 )
2 nvz.5 𝑍 = ( 0vec𝑈 )
3 nvz.6 𝑁 = ( normCV𝑈 )
4 eqid ( +𝑣𝑈 ) = ( +𝑣𝑈 )
5 eqid ( ·𝑠OLD𝑈 ) = ( ·𝑠OLD𝑈 )
6 1 4 5 2 3 nvi ( 𝑈 ∈ NrmCVec → ( ⟨ ( +𝑣𝑈 ) , ( ·𝑠OLD𝑈 ) ⟩ ∈ CVecOLD𝑁 : 𝑋 ⟶ ℝ ∧ ∀ 𝑥𝑋 ( ( ( 𝑁𝑥 ) = 0 → 𝑥 = 𝑍 ) ∧ ∀ 𝑦 ∈ ℂ ( 𝑁 ‘ ( 𝑦 ( ·𝑠OLD𝑈 ) 𝑥 ) ) = ( ( abs ‘ 𝑦 ) · ( 𝑁𝑥 ) ) ∧ ∀ 𝑦𝑋 ( 𝑁 ‘ ( 𝑥 ( +𝑣𝑈 ) 𝑦 ) ) ≤ ( ( 𝑁𝑥 ) + ( 𝑁𝑦 ) ) ) ) )
7 6 simp3d ( 𝑈 ∈ NrmCVec → ∀ 𝑥𝑋 ( ( ( 𝑁𝑥 ) = 0 → 𝑥 = 𝑍 ) ∧ ∀ 𝑦 ∈ ℂ ( 𝑁 ‘ ( 𝑦 ( ·𝑠OLD𝑈 ) 𝑥 ) ) = ( ( abs ‘ 𝑦 ) · ( 𝑁𝑥 ) ) ∧ ∀ 𝑦𝑋 ( 𝑁 ‘ ( 𝑥 ( +𝑣𝑈 ) 𝑦 ) ) ≤ ( ( 𝑁𝑥 ) + ( 𝑁𝑦 ) ) ) )
8 simp1 ( ( ( ( 𝑁𝑥 ) = 0 → 𝑥 = 𝑍 ) ∧ ∀ 𝑦 ∈ ℂ ( 𝑁 ‘ ( 𝑦 ( ·𝑠OLD𝑈 ) 𝑥 ) ) = ( ( abs ‘ 𝑦 ) · ( 𝑁𝑥 ) ) ∧ ∀ 𝑦𝑋 ( 𝑁 ‘ ( 𝑥 ( +𝑣𝑈 ) 𝑦 ) ) ≤ ( ( 𝑁𝑥 ) + ( 𝑁𝑦 ) ) ) → ( ( 𝑁𝑥 ) = 0 → 𝑥 = 𝑍 ) )
9 8 ralimi ( ∀ 𝑥𝑋 ( ( ( 𝑁𝑥 ) = 0 → 𝑥 = 𝑍 ) ∧ ∀ 𝑦 ∈ ℂ ( 𝑁 ‘ ( 𝑦 ( ·𝑠OLD𝑈 ) 𝑥 ) ) = ( ( abs ‘ 𝑦 ) · ( 𝑁𝑥 ) ) ∧ ∀ 𝑦𝑋 ( 𝑁 ‘ ( 𝑥 ( +𝑣𝑈 ) 𝑦 ) ) ≤ ( ( 𝑁𝑥 ) + ( 𝑁𝑦 ) ) ) → ∀ 𝑥𝑋 ( ( 𝑁𝑥 ) = 0 → 𝑥 = 𝑍 ) )
10 fveqeq2 ( 𝑥 = 𝐴 → ( ( 𝑁𝑥 ) = 0 ↔ ( 𝑁𝐴 ) = 0 ) )
11 eqeq1 ( 𝑥 = 𝐴 → ( 𝑥 = 𝑍𝐴 = 𝑍 ) )
12 10 11 imbi12d ( 𝑥 = 𝐴 → ( ( ( 𝑁𝑥 ) = 0 → 𝑥 = 𝑍 ) ↔ ( ( 𝑁𝐴 ) = 0 → 𝐴 = 𝑍 ) ) )
13 12 rspccv ( ∀ 𝑥𝑋 ( ( 𝑁𝑥 ) = 0 → 𝑥 = 𝑍 ) → ( 𝐴𝑋 → ( ( 𝑁𝐴 ) = 0 → 𝐴 = 𝑍 ) ) )
14 7 9 13 3syl ( 𝑈 ∈ NrmCVec → ( 𝐴𝑋 → ( ( 𝑁𝐴 ) = 0 → 𝐴 = 𝑍 ) ) )
15 14 imp ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( ( 𝑁𝐴 ) = 0 → 𝐴 = 𝑍 ) )
16 fveq2 ( 𝐴 = 𝑍 → ( 𝑁𝐴 ) = ( 𝑁𝑍 ) )
17 2 3 nvz0 ( 𝑈 ∈ NrmCVec → ( 𝑁𝑍 ) = 0 )
18 16 17 sylan9eqr ( ( 𝑈 ∈ NrmCVec ∧ 𝐴 = 𝑍 ) → ( 𝑁𝐴 ) = 0 )
19 18 ex ( 𝑈 ∈ NrmCVec → ( 𝐴 = 𝑍 → ( 𝑁𝐴 ) = 0 ) )
20 19 adantr ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( 𝐴 = 𝑍 → ( 𝑁𝐴 ) = 0 ) )
21 15 20 impbid ( ( 𝑈 ∈ NrmCVec ∧ 𝐴𝑋 ) → ( ( 𝑁𝐴 ) = 0 ↔ 𝐴 = 𝑍 ) )