Metamath Proof Explorer


Theorem rhmsubcrngclem1

Description: Lemma 1 for rhmsubcrngc . (Contributed by AV, 9-Mar-2020)

Ref Expression
Hypotheses rhmsubcrngc.c ⊢ C = RngCat ⁡ U
rhmsubcrngc.u ⊢ φ → U ∈ V
rhmsubcrngc.b ⊢ φ → B = Ring ∩ U
rhmsubcrngc.h ⊢ φ → H = RingHom ↾ B × B
Assertion rhmsubcrngclem1 ⊢ φ ∧ x ∈ B → Id ⁡ C ⁡ x ∈ x H x

Proof

Step Hyp Ref Expression
1 rhmsubcrngc.c ⊢ C = RngCat ⁡ U
2 rhmsubcrngc.u ⊢ φ → U ∈ V
3 rhmsubcrngc.b ⊢ φ → B = Ring ∩ U
4 rhmsubcrngc.h ⊢ φ → H = RingHom ↾ B × B
5 3 eleq2d ⊢ φ → x ∈ B ↔ x ∈ Ring ∩ U
6 elin ⊢ x ∈ Ring ∩ U ↔ x ∈ Ring ∧ x ∈ U
7 6 simplbi ⊢ x ∈ Ring ∩ U → x ∈ Ring
8 5 7 biimtrdi ⊢ φ → x ∈ B → x ∈ Ring
9 8 imp ⊢ φ ∧ x ∈ B → x ∈ Ring
10 eqid ⊢ Base x = Base x
11 10 idrhm ⊢ x ∈ Ring → I ↾ Base x ∈ x RingHom x
12 9 11 syl ⊢ φ ∧ x ∈ B → I ↾ Base x ∈ x RingHom x
13 eqid ⊢ Base C = Base C
14 eqid ⊢ Id ⁡ C = Id ⁡ C
15 2 adantr ⊢ φ ∧ x ∈ B → U ∈ V
16 ringrng ⊢ x ∈ Ring → x ∈ Rng
17 16 anim2i ⊢ x ∈ U ∧ x ∈ Ring → x ∈ U ∧ x ∈ Rng
18 17 ancoms ⊢ x ∈ Ring ∧ x ∈ U → x ∈ U ∧ x ∈ Rng
19 6 18 sylbi ⊢ x ∈ Ring ∩ U → x ∈ U ∧ x ∈ Rng
20 19 adantl ⊢ φ ∧ x ∈ Ring ∩ U → x ∈ U ∧ x ∈ Rng
21 elin ⊢ x ∈ U ∩ Rng ↔ x ∈ U ∧ x ∈ Rng
22 20 21 sylibr ⊢ φ ∧ x ∈ Ring ∩ U → x ∈ U ∩ Rng
23 1 13 2 rngcbas ⊢ φ → Base C = U ∩ Rng
24 23 adantr ⊢ φ ∧ x ∈ Ring ∩ U → Base C = U ∩ Rng
25 22 24 eleqtrrd ⊢ φ ∧ x ∈ Ring ∩ U → x ∈ Base C
26 25 ex ⊢ φ → x ∈ Ring ∩ U → x ∈ Base C
27 5 26 sylbid ⊢ φ → x ∈ B → x ∈ Base C
28 27 imp ⊢ φ ∧ x ∈ B → x ∈ Base C
29 1 13 14 15 28 10 rngcid ⊢ φ ∧ x ∈ B → Id ⁡ C ⁡ x = I ↾ Base x
30 4 oveqdr ⊢ φ ∧ x ∈ B → x H x = x RingHom ↾ B × B x
31 eqid ⊢ RingCat ⁡ U = RingCat ⁡ U
32 eqid ⊢ Base RingCat ⁡ U = Base RingCat ⁡ U
33 eqid ⊢ Hom ⁡ RingCat ⁡ U = Hom ⁡ RingCat ⁡ U
34 31 32 2 33 ringchomfval ⊢ φ → Hom ⁡ RingCat ⁡ U = RingHom ↾ Base RingCat ⁡ U × Base RingCat ⁡ U
35 31 32 2 ringcbas ⊢ φ → Base RingCat ⁡ U = U ∩ Ring
36 incom ⊢ Ring ∩ U = U ∩ Ring
37 3 36 eqtrdi ⊢ φ → B = U ∩ Ring
38 37 eqcomd ⊢ φ → U ∩ Ring = B
39 35 38 eqtrd ⊢ φ → Base RingCat ⁡ U = B
40 39 sqxpeqd ⊢ φ → Base RingCat ⁡ U × Base RingCat ⁡ U = B × B
41 40 reseq2d ⊢ φ → RingHom ↾ Base RingCat ⁡ U × Base RingCat ⁡ U = RingHom ↾ B × B
42 34 41 eqtrd ⊢ φ → Hom ⁡ RingCat ⁡ U = RingHom ↾ B × B
43 42 adantr ⊢ φ ∧ x ∈ B → Hom ⁡ RingCat ⁡ U = RingHom ↾ B × B
44 43 eqcomd ⊢ φ ∧ x ∈ B → RingHom ↾ B × B = Hom ⁡ RingCat ⁡ U
45 44 oveqd ⊢ φ ∧ x ∈ B → x RingHom ↾ B × B x = x Hom ⁡ RingCat ⁡ U x
46 37 eleq2d ⊢ φ → x ∈ B ↔ x ∈ U ∩ Ring
47 46 biimpa ⊢ φ ∧ x ∈ B → x ∈ U ∩ Ring
48 35 adantr ⊢ φ ∧ x ∈ B → Base RingCat ⁡ U = U ∩ Ring
49 47 48 eleqtrrd ⊢ φ ∧ x ∈ B → x ∈ Base RingCat ⁡ U
50 31 32 15 33 49 49 ringchom ⊢ φ ∧ x ∈ B → x Hom ⁡ RingCat ⁡ U x = x RingHom x
51 30 45 50 3eqtrd ⊢ φ ∧ x ∈ B → x H x = x RingHom x
52 12 29 51 3eltr4d ⊢ φ ∧ x ∈ B → Id ⁡ C ⁡ x ∈ x H x