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 )