Step |
Hyp |
Ref |
Expression |
1 |
|
erng0g.b |
⊢ 𝐵 = ( Base ‘ 𝐾 ) |
2 |
|
erng0g.h |
⊢ 𝐻 = ( LHyp ‘ 𝐾 ) |
3 |
|
erng0g.t |
⊢ 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 ) |
4 |
|
erng0g.d |
⊢ 𝐷 = ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 ) |
5 |
|
erng0g.o |
⊢ 𝑂 = ( 𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵 ) ) |
6 |
|
erng0g.z |
⊢ 0 = ( 0g ‘ 𝐷 ) |
7 |
|
eqid |
⊢ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) |
8 |
|
eqid |
⊢ ( +g ‘ 𝐷 ) = ( +g ‘ 𝐷 ) |
9 |
2 3 7 4 8
|
erngfplus |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( +g ‘ 𝐷 ) = ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑠 ‘ 𝑓 ) ∘ ( 𝑡 ‘ 𝑓 ) ) ) ) ) |
10 |
9
|
oveqd |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( 𝑂 ( +g ‘ 𝐷 ) 𝑂 ) = ( 𝑂 ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑠 ‘ 𝑓 ) ∘ ( 𝑡 ‘ 𝑓 ) ) ) ) 𝑂 ) ) |
11 |
1 2 3 7 5
|
tendo0cl |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝑂 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ) |
12 |
|
eqid |
⊢ ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑠 ‘ 𝑓 ) ∘ ( 𝑡 ‘ 𝑓 ) ) ) ) = ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑠 ‘ 𝑓 ) ∘ ( 𝑡 ‘ 𝑓 ) ) ) ) |
13 |
1 2 3 7 5 12
|
tendo0pl |
⊢ ( ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) ∧ 𝑂 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ) → ( 𝑂 ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑠 ‘ 𝑓 ) ∘ ( 𝑡 ‘ 𝑓 ) ) ) ) 𝑂 ) = 𝑂 ) |
14 |
11 13
|
mpdan |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( 𝑂 ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓 ∈ 𝑇 ↦ ( ( 𝑠 ‘ 𝑓 ) ∘ ( 𝑡 ‘ 𝑓 ) ) ) ) 𝑂 ) = 𝑂 ) |
15 |
10 14
|
eqtrd |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( 𝑂 ( +g ‘ 𝐷 ) 𝑂 ) = 𝑂 ) |
16 |
2 4
|
eringring |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝐷 ∈ Ring ) |
17 |
|
ringgrp |
⊢ ( 𝐷 ∈ Ring → 𝐷 ∈ Grp ) |
18 |
16 17
|
syl |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝐷 ∈ Grp ) |
19 |
|
eqid |
⊢ ( Base ‘ 𝐷 ) = ( Base ‘ 𝐷 ) |
20 |
2 3 7 4 19
|
erngbase |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( Base ‘ 𝐷 ) = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ) |
21 |
11 20
|
eleqtrrd |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 𝑂 ∈ ( Base ‘ 𝐷 ) ) |
22 |
19 8 6
|
grpid |
⊢ ( ( 𝐷 ∈ Grp ∧ 𝑂 ∈ ( Base ‘ 𝐷 ) ) → ( ( 𝑂 ( +g ‘ 𝐷 ) 𝑂 ) = 𝑂 ↔ 0 = 𝑂 ) ) |
23 |
18 21 22
|
syl2anc |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → ( ( 𝑂 ( +g ‘ 𝐷 ) 𝑂 ) = 𝑂 ↔ 0 = 𝑂 ) ) |
24 |
15 23
|
mpbid |
⊢ ( ( 𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ) → 0 = 𝑂 ) |