Metamath Proof Explorer


Theorem dihsslss

Description: The isomorphism H maps to subspaces. (Contributed by NM, 14-Mar-2014)

Ref Expression
Hypotheses dihsslss.h 𝐻 = ( LHyp ‘ 𝐾 )
dihsslss.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
dihsslss.i 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
dihsslss.s 𝑆 = ( LSubSp ‘ 𝑈 )
Assertion dihsslss ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ran 𝐼𝑆 )

Proof

Step Hyp Ref Expression
1 dihsslss.h 𝐻 = ( LHyp ‘ 𝐾 )
2 dihsslss.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
3 dihsslss.i 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
4 dihsslss.s 𝑆 = ( LSubSp ‘ 𝑈 )
5 1 3 dihcnvid2 ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑥 ∈ ran 𝐼 ) → ( 𝐼 ‘ ( 𝐼𝑥 ) ) = 𝑥 )
6 eqid ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
7 6 1 3 dihcnvcl ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑥 ∈ ran 𝐼 ) → ( 𝐼𝑥 ) ∈ ( Base ‘ 𝐾 ) )
8 6 1 3 2 4 dihlss ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ ( 𝐼𝑥 ) ∈ ( Base ‘ 𝐾 ) ) → ( 𝐼 ‘ ( 𝐼𝑥 ) ) ∈ 𝑆 )
9 7 8 syldan ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑥 ∈ ran 𝐼 ) → ( 𝐼 ‘ ( 𝐼𝑥 ) ) ∈ 𝑆 )
10 5 9 eqeltrrd ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑥 ∈ ran 𝐼 ) → 𝑥𝑆 )
11 10 ex ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝑥 ∈ ran 𝐼𝑥𝑆 ) )
12 11 ssrdv ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ran 𝐼𝑆 )