Metamath Proof Explorer


Theorem lsslinds

Description: Linear independence is unchanged by working in a subspace. (Contributed by Stefan O'Rear, 24-Feb-2015)

Ref Expression
Hypotheses lsslindf.u 𝑈 = ( LSubSp ‘ 𝑊 )
lsslindf.x 𝑋 = ( 𝑊s 𝑆 )
Assertion lsslinds ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( 𝐹 ∈ ( LIndS ‘ 𝑋 ) ↔ 𝐹 ∈ ( LIndS ‘ 𝑊 ) ) )

Proof

Step Hyp Ref Expression
1 lsslindf.u 𝑈 = ( LSubSp ‘ 𝑊 )
2 lsslindf.x 𝑋 = ( 𝑊s 𝑆 )
3 eqid ( Base ‘ 𝑊 ) = ( Base ‘ 𝑊 )
4 3 1 lssss ( 𝑆𝑈𝑆 ⊆ ( Base ‘ 𝑊 ) )
5 2 3 ressbas2 ( 𝑆 ⊆ ( Base ‘ 𝑊 ) → 𝑆 = ( Base ‘ 𝑋 ) )
6 4 5 syl ( 𝑆𝑈𝑆 = ( Base ‘ 𝑋 ) )
7 6 3ad2ant2 ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → 𝑆 = ( Base ‘ 𝑋 ) )
8 7 sseq2d ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( 𝐹𝑆𝐹 ⊆ ( Base ‘ 𝑋 ) ) )
9 4 3ad2ant2 ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → 𝑆 ⊆ ( Base ‘ 𝑊 ) )
10 sstr2 ( 𝐹𝑆 → ( 𝑆 ⊆ ( Base ‘ 𝑊 ) → 𝐹 ⊆ ( Base ‘ 𝑊 ) ) )
11 9 10 mpan9 ( ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) ∧ 𝐹𝑆 ) → 𝐹 ⊆ ( Base ‘ 𝑊 ) )
12 simpl3 ( ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) ∧ 𝐹 ⊆ ( Base ‘ 𝑊 ) ) → 𝐹𝑆 )
13 11 12 impbida ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( 𝐹𝑆𝐹 ⊆ ( Base ‘ 𝑊 ) ) )
14 8 13 bitr3d ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( 𝐹 ⊆ ( Base ‘ 𝑋 ) ↔ 𝐹 ⊆ ( Base ‘ 𝑊 ) ) )
15 rnresi ran ( I ↾ 𝐹 ) = 𝐹
16 15 sseq1i ( ran ( I ↾ 𝐹 ) ⊆ 𝑆𝐹𝑆 )
17 1 2 lsslindf ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈 ∧ ran ( I ↾ 𝐹 ) ⊆ 𝑆 ) → ( ( I ↾ 𝐹 ) LIndF 𝑋 ↔ ( I ↾ 𝐹 ) LIndF 𝑊 ) )
18 16 17 syl3an3br ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( ( I ↾ 𝐹 ) LIndF 𝑋 ↔ ( I ↾ 𝐹 ) LIndF 𝑊 ) )
19 14 18 anbi12d ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( ( 𝐹 ⊆ ( Base ‘ 𝑋 ) ∧ ( I ↾ 𝐹 ) LIndF 𝑋 ) ↔ ( 𝐹 ⊆ ( Base ‘ 𝑊 ) ∧ ( I ↾ 𝐹 ) LIndF 𝑊 ) ) )
20 2 ovexi 𝑋 ∈ V
21 eqid ( Base ‘ 𝑋 ) = ( Base ‘ 𝑋 )
22 21 islinds ( 𝑋 ∈ V → ( 𝐹 ∈ ( LIndS ‘ 𝑋 ) ↔ ( 𝐹 ⊆ ( Base ‘ 𝑋 ) ∧ ( I ↾ 𝐹 ) LIndF 𝑋 ) ) )
23 20 22 mp1i ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( 𝐹 ∈ ( LIndS ‘ 𝑋 ) ↔ ( 𝐹 ⊆ ( Base ‘ 𝑋 ) ∧ ( I ↾ 𝐹 ) LIndF 𝑋 ) ) )
24 3 islinds ( 𝑊 ∈ LMod → ( 𝐹 ∈ ( LIndS ‘ 𝑊 ) ↔ ( 𝐹 ⊆ ( Base ‘ 𝑊 ) ∧ ( I ↾ 𝐹 ) LIndF 𝑊 ) ) )
25 24 3ad2ant1 ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( 𝐹 ∈ ( LIndS ‘ 𝑊 ) ↔ ( 𝐹 ⊆ ( Base ‘ 𝑊 ) ∧ ( I ↾ 𝐹 ) LIndF 𝑊 ) ) )
26 19 23 25 3bitr4d ( ( 𝑊 ∈ LMod ∧ 𝑆𝑈𝐹𝑆 ) → ( 𝐹 ∈ ( LIndS ‘ 𝑋 ) ↔ 𝐹 ∈ ( LIndS ‘ 𝑊 ) ) )