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Metamath Proof Explorer

Theorem dihordlem6

Description: Part of proof of Lemma N of Crawley p. 122 line 35. (Contributed by NM, 3-Mar-2014)

Ref Expression
Hypotheses dihordlem8.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
dihordlem8.l ⊢ ≤ = ( le ‘ 𝐾 )
dihordlem8.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
dihordlem8.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
dihordlem8.p ⊢ 𝑃 = ( ( oc ‘ 𝐾 ) ‘ 𝑊 )
dihordlem8.o ⊢ 𝑂 = ( ℎ ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
dihordlem8.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
dihordlem8.e ⊢ 𝐸 = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 )
dihordlem8.u ⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
dihordlem8.s ⊢ + = ( +g ‘ 𝑈 )
dihordlem8.g ⊢ 𝐺 = ( ℩ ℎ ∈ 𝑇 ( ℎ ‘ 𝑃 ) = 𝑅 )
Assertion dihordlem6 ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ∧ ( 𝑅 ∈ 𝐴 ∧ ¬ 𝑅 ≤ 𝑊 ) ) ∧ ( 𝑠 ∈ 𝐸 ∧ 𝑔 ∈ 𝑇 ) ) → ( ⟨ ( 𝑠 ‘ 𝐺 ) , 𝑠 ⟩ + ⟨ 𝑔 , 𝑂 ⟩ ) = ⟨ ( ( 𝑠 ‘ 𝐺 ) ∘ 𝑔 ) , 𝑠 ⟩ )

Proof

Step Hyp Ref Expression
1 dihordlem8.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
2 dihordlem8.l ⊢ ≤ = ( le ‘ 𝐾 )
3 dihordlem8.a ⊢ 𝐴 = ( Atoms ‘ 𝐾 )
4 dihordlem8.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
5 dihordlem8.p ⊢ 𝑃 = ( ( oc ‘ 𝐾 ) ‘ 𝑊 )
6 dihordlem8.o ⊢ 𝑂 = ( ℎ ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
7 dihordlem8.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
8 dihordlem8.e ⊢ 𝐸 = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 )
9 dihordlem8.u ⊢ 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
10 dihordlem8.s ⊢ + = ( +g ‘ 𝑈 )
11 dihordlem8.g ⊢ 𝐺 = ( ℩ ℎ ∈ 𝑇 ( ℎ ‘ 𝑃 ) = 𝑅 )
12 simp1 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ∧ ( 𝑅 ∈ 𝐴 ∧ ¬ 𝑅 ≤ 𝑊 ) ) ∧ ( 𝑠 ∈ 𝐸 ∧ 𝑔 ∈ 𝑇 ) ) → ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) )
13 simp2r ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ∧ ( 𝑅 ∈ 𝐴 ∧ ¬ 𝑅 ≤ 𝑊 ) ) ∧ ( 𝑠 ∈ 𝐸 ∧ 𝑔 ∈ 𝑇 ) ) → ( 𝑅 ∈ 𝐴 ∧ ¬ 𝑅 ≤ 𝑊 ) )
14 simp2l ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ∧ ( 𝑅 ∈ 𝐴 ∧ ¬ 𝑅 ≤ 𝑊 ) ) ∧ ( 𝑠 ∈ 𝐸 ∧ 𝑔 ∈ 𝑇 ) ) → ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) )
15 simp3 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ∧ ( 𝑅 ∈ 𝐴 ∧ ¬ 𝑅 ≤ 𝑊 ) ) ∧ ( 𝑠 ∈ 𝐸 ∧ 𝑔 ∈ 𝑇 ) ) → ( 𝑠 ∈ 𝐸 ∧ 𝑔 ∈ 𝑇 ) )
16 1 2 3 4 5 6 7 8 9 10 11 cdlemn6 ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( ( 𝑅 ∈ 𝐴 ∧ ¬ 𝑅 ≤ 𝑊 ) ∧ ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ) ∧ ( 𝑠 ∈ 𝐸 ∧ 𝑔 ∈ 𝑇 ) ) → ( ⟨ ( 𝑠 ‘ 𝐺 ) , 𝑠 ⟩ + ⟨ 𝑔 , 𝑂 ⟩ ) = ⟨ ( ( 𝑠 ‘ 𝐺 ) ∘ 𝑔 ) , 𝑠 ⟩ )
17 12 13 14 15 16 syl121anc ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( ( 𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊 ) ∧ ( 𝑅 ∈ 𝐴 ∧ ¬ 𝑅 ≤ 𝑊 ) ) ∧ ( 𝑠 ∈ 𝐸 ∧ 𝑔 ∈ 𝑇 ) ) → ( ⟨ ( 𝑠 ‘ 𝐺 ) , 𝑠 ⟩ + ⟨ 𝑔 , 𝑂 ⟩ ) = ⟨ ( ( 𝑠 ‘ 𝐺 ) ∘ 𝑔 ) , 𝑠 ⟩ )