Metamath Proof Explorer


Theorem h1dn0

Description: A nonzero vector generates a (nonzero) 1-dimensional subspace. (Contributed by NM, 22-Jul-2001) (New usage is discouraged.)

Ref Expression
Assertion h1dn0 ( ( 𝐴 ∈ ℋ ∧ 𝐴 ≠ 0ℎ ) → ( ⊥ ‘ ( ⊥ ‘ { 𝐴 } ) ) ≠ 0ℋ )

Proof

Step Hyp Ref Expression
1 h1did ⊢ ( 𝐴 ∈ ℋ → 𝐴 ∈ ( ⊥ ‘ ( ⊥ ‘ { 𝐴 } ) ) )
2 eleq2 ⊢ ( ( ⊥ ‘ ( ⊥ ‘ { 𝐴 } ) ) = 0ℋ → ( 𝐴 ∈ ( ⊥ ‘ ( ⊥ ‘ { 𝐴 } ) ) ↔ 𝐴 ∈ 0ℋ ) )
3 1 2 syl5ibcom ⊢ ( 𝐴 ∈ ℋ → ( ( ⊥ ‘ ( ⊥ ‘ { 𝐴 } ) ) = 0ℋ → 𝐴 ∈ 0ℋ ) )
4 elch0 ⊢ ( 𝐴 ∈ 0ℋ ↔ 𝐴 = 0ℎ )
5 3 4 imbitrdi ⊢ ( 𝐴 ∈ ℋ → ( ( ⊥ ‘ ( ⊥ ‘ { 𝐴 } ) ) = 0ℋ → 𝐴 = 0ℎ ) )
6 5 necon3d ⊢ ( 𝐴 ∈ ℋ → ( 𝐴 ≠ 0ℎ → ( ⊥ ‘ ( ⊥ ‘ { 𝐴 } ) ) ≠ 0ℋ ) )
7 6 imp ⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐴 ≠ 0ℎ ) → ( ⊥ ‘ ( ⊥ ‘ { 𝐴 } ) ) ≠ 0ℋ )