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Metamath Proof Explorer

Theorem erngdvlem1

Description: Lemma for eringring . (Contributed by NM, 4-Aug-2013)

Ref Expression
Hypotheses ernggrp.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
ernggrp.d ⊢ 𝐷 = ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 )
erngdv.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
erngdv.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
erngdv.e ⊢ 𝐸 = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 )
erngdv.p ⊢ 𝑃 = ( 𝑎 ∈ 𝐸 , 𝑏 ∈ 𝐸 ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑎 ‘ 𝑓 ) ∘ ( 𝑏 ‘ 𝑓 ) ) ) )
erngdv.o ⊢ 0 = ( 𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
erngdv.i ⊢ 𝐼 = ( 𝑎 ∈ 𝐸 ↦ ( 𝑓 ∈ 𝑇 ↦ ◡ ( 𝑎 ‘ 𝑓 ) ) )
Assertion erngdvlem1 ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝐷 ∈ Grp )

Proof

Step Hyp Ref Expression
1 ernggrp.h ⊢ 𝐻 = ( LHyp ‘ 𝐾 )
2 ernggrp.d ⊢ 𝐷 = ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 )
3 erngdv.b ⊢ 𝐵 = ( Base ‘ 𝐾 )
4 erngdv.t ⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
5 erngdv.e ⊢ 𝐸 = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 )
6 erngdv.p ⊢ 𝑃 = ( 𝑎 ∈ 𝐸 , 𝑏 ∈ 𝐸 ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑎 ‘ 𝑓 ) ∘ ( 𝑏 ‘ 𝑓 ) ) ) )
7 erngdv.o ⊢ 0 = ( 𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵 ) )
8 erngdv.i ⊢ 𝐼 = ( 𝑎 ∈ 𝐸 ↦ ( 𝑓 ∈ 𝑇 ↦ ◡ ( 𝑎 ‘ 𝑓 ) ) )
9 eqid ⊢ ( Base ‘ 𝐷 ) = ( Base ‘ 𝐷 )
10 1 4 5 2 9 erngbase ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( Base ‘ 𝐷 ) = 𝐸 )
11 10 eqcomd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝐸 = ( Base ‘ 𝐷 ) )
12 eqid ⊢ ( +g ‘ 𝐷 ) = ( +g ‘ 𝐷 )
13 1 4 5 2 12 erngfplus ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( +g ‘ 𝐷 ) = ( 𝑎 ∈ 𝐸 , 𝑏 ∈ 𝐸 ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑎 ‘ 𝑓 ) ∘ ( 𝑏 ‘ 𝑓 ) ) ) ) )
14 6 13 eqtr4id ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝑃 = ( +g ‘ 𝐷 ) )
15 1 4 5 6 tendoplcl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑠 ∈ 𝐸 ∧ 𝑡 ∈ 𝐸 ) → ( 𝑠 𝑃 𝑡 ) ∈ 𝐸 )
16 1 4 5 6 tendoplass ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ ( 𝑠 ∈ 𝐸 ∧ 𝑡 ∈ 𝐸 ∧ 𝑢 ∈ 𝐸 ) ) → ( ( 𝑠 𝑃 𝑡 ) 𝑃 𝑢 ) = ( 𝑠 𝑃 ( 𝑡 𝑃 𝑢 ) ) )
17 3 1 4 5 7 tendo0cl ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 0 ∈ 𝐸 )
18 3 1 4 5 7 6 tendo0pl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑠 ∈ 𝐸 ) → ( 0 𝑃 𝑠 ) = 𝑠 )
19 1 4 5 8 tendoicl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑠 ∈ 𝐸 ) → ( 𝐼 ‘ 𝑠 ) ∈ 𝐸 )
20 1 4 5 8 3 6 7 tendoipl ⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑠 ∈ 𝐸 ) → ( ( 𝐼 ‘ 𝑠 ) 𝑃 𝑠 ) = 0 )
21 11 14 15 16 17 18 19 20 isgrpd ⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝐷 ∈ Grp )