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Metamath Proof Explorer

Theorem dih1dimc

Description: Isomorphism H at an atom not under W . (Contributed by NM, 27-Apr-2014)

Ref Expression
Hypotheses dih1dimc.l ⊢ ≤ = ( le ‘ 𝐾 )
dih1dimc.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
dih1dimc.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
dih1dimc.p ⊢ 𝑃 = ( ( oc ‘ 𝐾 ) ‘ 𝑊 )
dih1dimc.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
dih1dimc.i ⊢ 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
dih1dimc.u ⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
dih1dimc.n ⊢ 𝑁 = ( LSpan ‘ 𝑈 )
dih1dimc.f ⊢ 𝐹 = ( ℩ 𝑓 ∈ 𝑇 ( 𝑓 ‘ 𝑃 ) = 𝑄 )
Assertion dih1dimc ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) → ( 𝐼 ‘ 𝑄 ) = ( 𝑁 ‘ { ⟨ 𝐹 , ( I ↾ 𝑇 ) ⟩ } ) )

Proof

Step Hyp Ref Expression
1 dih1dimc.l ⊢ ≤ = ( le ‘ 𝐾 )
2 dih1dimc.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
3 dih1dimc.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
4 dih1dimc.p ⊢ 𝑃 = ( ( oc ‘ 𝐾 ) ‘ 𝑊 )
5 dih1dimc.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
6 dih1dimc.i ⊢ 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
7 dih1dimc.u ⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
8 dih1dimc.n ⊢ 𝑁 = ( LSpan ‘ 𝑈 )
9 dih1dimc.f ⊢ 𝐹 = ( ℩ 𝑓 ∈ 𝑇 ( 𝑓 ‘ 𝑃 ) = 𝑄 )
10 eqid ⊢ ( ( DIsoC ‘ 𝐾 ) ‘ 𝑊 ) = ( ( DIsoC ‘ 𝐾 ) ‘ 𝑊 )
11 1 2 3 10 6 dihvalcqat ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) → ( 𝐼 ‘ 𝑄 ) = ( ( ( DIsoC ‘ 𝐾 ) ‘ 𝑊 ) ‘ 𝑄 ) )
12 1 2 3 4 5 10 7 8 9 diclspsn ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) → ( ( ( DIsoC ‘ 𝐾 ) ‘ 𝑊 ) ‘ 𝑄 ) = ( 𝑁 ‘ { ⟨ 𝐹 , ( I ↾ 𝑇 ) ⟩ } ) )
13 11 12 eqtrd ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) → ( 𝐼 ‘ 𝑄 ) = ( 𝑁 ‘ { ⟨ 𝐹 , ( I ↾ 𝑇 ) ⟩ } ) )