lbsind

Description: A basis is linearly independent; that is, every element has a span which trivially intersects the span of the remainder of the basis. (Contributed by Mario Carneiro, 12-Jan-2015)

	Ref	Expression
Hypotheses	lbsss.v	⊢ 𝑉 = ( Base ‘ 𝑊 )
	lbsss.j	⊢ 𝐽 = ( LBasis ‘ 𝑊 )
	lbssp.n	⊢ 𝑁 = ( LSpan ‘ 𝑊 )
	lbsind.f	⊢ 𝐹 = ( Scalar ‘ 𝑊 )
	lbsind.s	⊢ · = ( ·_𝑠 ‘ 𝑊 )
	lbsind.k	⊢ 𝐾 = ( Base ‘ 𝐹 )
	lbsind.z	⊢ 0 = ( 0_g ‘ 𝐹 )
Assertion	lbsind	⊢ ( ( ( 𝐵 ∈ 𝐽 ∧ 𝐸 ∈ 𝐵 ) ∧ ( 𝐴 ∈ 𝐾 ∧ 𝐴 ≠ 0 ) ) → ¬ ( 𝐴 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) )

Proof

Step	Hyp	Ref	Expression
1		lbsss.v	⊢ 𝑉 = ( Base ‘ 𝑊 )
2		lbsss.j	⊢ 𝐽 = ( LBasis ‘ 𝑊 )
3		lbssp.n	⊢ 𝑁 = ( LSpan ‘ 𝑊 )
4		lbsind.f	⊢ 𝐹 = ( Scalar ‘ 𝑊 )
5		lbsind.s	⊢ · = ( ·_𝑠 ‘ 𝑊 )
6		lbsind.k	⊢ 𝐾 = ( Base ‘ 𝐹 )
7		lbsind.z	⊢ 0 = ( 0_g ‘ 𝐹 )
8		eldifsn	⊢ ( 𝐴 ∈ ( 𝐾 ∖ { 0 } ) ↔ ( 𝐴 ∈ 𝐾 ∧ 𝐴 ≠ 0 ) )
9		elfvdm	⊢ ( 𝐵 ∈ ( LBasis ‘ 𝑊 ) → 𝑊 ∈ dom LBasis )
10	9 2	eleq2s	⊢ ( 𝐵 ∈ 𝐽 → 𝑊 ∈ dom LBasis )
11	1 4 5 6 2 3 7	islbs	⊢ ( 𝑊 ∈ dom LBasis → ( 𝐵 ∈ 𝐽 ↔ ( 𝐵 ⊆ 𝑉 ∧ ( 𝑁 ‘ 𝐵 ) = 𝑉 ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ ( 𝐾 ∖ { 0 } ) ¬ ( 𝑦 · 𝑥 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝑥 } ) ) ) ) )
12	10 11	syl	⊢ ( 𝐵 ∈ 𝐽 → ( 𝐵 ∈ 𝐽 ↔ ( 𝐵 ⊆ 𝑉 ∧ ( 𝑁 ‘ 𝐵 ) = 𝑉 ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ ( 𝐾 ∖ { 0 } ) ¬ ( 𝑦 · 𝑥 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝑥 } ) ) ) ) )
13	12	ibi	⊢ ( 𝐵 ∈ 𝐽 → ( 𝐵 ⊆ 𝑉 ∧ ( 𝑁 ‘ 𝐵 ) = 𝑉 ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ ( 𝐾 ∖ { 0 } ) ¬ ( 𝑦 · 𝑥 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝑥 } ) ) ) )
14	13	simp3d	⊢ ( 𝐵 ∈ 𝐽 → ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ ( 𝐾 ∖ { 0 } ) ¬ ( 𝑦 · 𝑥 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝑥 } ) ) )
15		oveq2	⊢ ( 𝑥 = 𝐸 → ( 𝑦 · 𝑥 ) = ( 𝑦 · 𝐸 ) )
16		sneq	⊢ ( 𝑥 = 𝐸 → { 𝑥 } = { 𝐸 } )
17	16	difeq2d	⊢ ( 𝑥 = 𝐸 → ( 𝐵 ∖ { 𝑥 } ) = ( 𝐵 ∖ { 𝐸 } ) )
18	17	fveq2d	⊢ ( 𝑥 = 𝐸 → ( 𝑁 ‘ ( 𝐵 ∖ { 𝑥 } ) ) = ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) )
19	15 18	eleq12d	⊢ ( 𝑥 = 𝐸 → ( ( 𝑦 · 𝑥 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝑥 } ) ) ↔ ( 𝑦 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) ) )
20	19	notbid	⊢ ( 𝑥 = 𝐸 → ( ¬ ( 𝑦 · 𝑥 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝑥 } ) ) ↔ ¬ ( 𝑦 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) ) )
21		oveq1	⊢ ( 𝑦 = 𝐴 → ( 𝑦 · 𝐸 ) = ( 𝐴 · 𝐸 ) )
22	21	eleq1d	⊢ ( 𝑦 = 𝐴 → ( ( 𝑦 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) ↔ ( 𝐴 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) ) )
23	22	notbid	⊢ ( 𝑦 = 𝐴 → ( ¬ ( 𝑦 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) ↔ ¬ ( 𝐴 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) ) )
24	20 23	rspc2v	⊢ ( ( 𝐸 ∈ 𝐵 ∧ 𝐴 ∈ ( 𝐾 ∖ { 0 } ) ) → ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ ( 𝐾 ∖ { 0 } ) ¬ ( 𝑦 · 𝑥 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝑥 } ) ) → ¬ ( 𝐴 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) ) )
25	14 24	syl5com	⊢ ( 𝐵 ∈ 𝐽 → ( ( 𝐸 ∈ 𝐵 ∧ 𝐴 ∈ ( 𝐾 ∖ { 0 } ) ) → ¬ ( 𝐴 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) ) )
26	25	impl	⊢ ( ( ( 𝐵 ∈ 𝐽 ∧ 𝐸 ∈ 𝐵 ) ∧ 𝐴 ∈ ( 𝐾 ∖ { 0 } ) ) → ¬ ( 𝐴 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) )
27	8 26	sylan2br	⊢ ( ( ( 𝐵 ∈ 𝐽 ∧ 𝐸 ∈ 𝐵 ) ∧ ( 𝐴 ∈ 𝐾 ∧ 𝐴 ≠ 0 ) ) → ¬ ( 𝐴 · 𝐸 ) ∈ ( 𝑁 ‘ ( 𝐵 ∖ { 𝐸 } ) ) )

Metamath Proof Explorer

Theorem lbsind

Proof