Metamath Proof Explorer


Theorem rhmsubclem3

Description: Lemma 3 for rhmsubc . (Contributed by AV, 2-Mar-2020)

Ref Expression
Hypotheses rngcrescrhm.u ⊢ φ → U ∈ V
rngcrescrhm.c ⊢ C = RngCat ⁡ U
rngcrescrhm.r ⊢ φ → R = Ring ∩ U
rngcrescrhm.h ⊢ H = RingHom ↾ R × R
Assertion rhmsubclem3 ⊢ φ ∧ x ∈ R → Id ⁡ RngCat ⁡ U ⁡ x ∈ x H x

Proof

Step Hyp Ref Expression
1 rngcrescrhm.u ⊢ φ → U ∈ V
2 rngcrescrhm.c ⊢ C = RngCat ⁡ U
3 rngcrescrhm.r ⊢ φ → R = Ring ∩ U
4 rngcrescrhm.h ⊢ H = RingHom ↾ R × R
5 3 eleq2d ⊢ φ → x ∈ R ↔ x ∈ Ring ∩ U
6 elinel1 ⊢ x ∈ Ring ∩ U → x ∈ Ring
7 5 6 biimtrdi ⊢ φ → x ∈ R → x ∈ Ring
8 7 imp ⊢ φ ∧ x ∈ R → x ∈ Ring
9 eqid ⊢ Base x = Base x
10 9 idrhm ⊢ x ∈ Ring → I ↾ Base x ∈ x RingHom x
11 8 10 syl ⊢ φ ∧ x ∈ R → I ↾ Base x ∈ x RingHom x
12 eqid ⊢ Base C = Base C
13 2 eqcomi ⊢ RngCat ⁡ U = C
14 13 fveq2i ⊢ Id ⁡ RngCat ⁡ U = Id ⁡ C
15 1 adantr ⊢ φ ∧ x ∈ R → U ∈ V
16 incom ⊢ Ring ∩ U = U ∩ Ring
17 ringssrng ⊢ Ring ⊆ Rng
18 sslin ⊢ Ring ⊆ Rng → U ∩ Ring ⊆ U ∩ Rng
19 17 18 mp1i ⊢ φ → U ∩ Ring ⊆ U ∩ Rng
20 16 19 eqsstrid ⊢ φ → Ring ∩ U ⊆ U ∩ Rng
21 2 12 1 rngcbas ⊢ φ → Base C = U ∩ Rng
22 20 3 21 3sstr4d ⊢ φ → R ⊆ Base C
23 22 sselda ⊢ φ ∧ x ∈ R → x ∈ Base C
24 2 12 14 15 23 9 rngcid ⊢ φ ∧ x ∈ R → Id ⁡ RngCat ⁡ U ⁡ x = I ↾ Base x
25 1 2 3 4 rhmsubclem2 ⊢ φ ∧ x ∈ R ∧ x ∈ R → x H x = x RingHom x
26 25 3anidm23 ⊢ φ ∧ x ∈ R → x H x = x RingHom x
27 11 24 26 3eltr4d ⊢ φ ∧ x ∈ R → Id ⁡ RngCat ⁡ U ⁡ x ∈ x H x