Metamath Proof Explorer


Theorem rnghmsubcsetclem1

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

Ref Expression
Hypotheses rnghmsubcsetc.c C = ExtStrCat U
rnghmsubcsetc.u φ U V
rnghmsubcsetc.b φ B = Rng U
rnghmsubcsetc.h φ H = RngHomo B × B
Assertion rnghmsubcsetclem1 φ x B Id C x x H x

Proof

Step Hyp Ref Expression
1 rnghmsubcsetc.c C = ExtStrCat U
2 rnghmsubcsetc.u φ U V
3 rnghmsubcsetc.b φ B = Rng U
4 rnghmsubcsetc.h φ H = RngHomo B × B
5 3 eleq2d φ x B x Rng U
6 elin x Rng U x Rng x U
7 6 simplbi x Rng U x Rng
8 5 7 syl6bi φ x B x Rng
9 8 imp φ x B x Rng
10 eqid Base x = Base x
11 10 idrnghm x Rng I Base x x RngHomo x
12 9 11 syl φ x B I Base x x RngHomo x
13 eqid Id C = Id C
14 2 adantr φ x B U V
15 6 simprbi x Rng U x U
16 5 15 syl6bi φ x B x U
17 16 imp φ x B x U
18 1 13 14 17 estrcid φ x B Id C x = I Base x
19 4 oveqdr φ x B x H x = x RngHomo B × B x
20 eqid RngCat U = RngCat U
21 eqid Base RngCat U = Base RngCat U
22 eqid Hom RngCat U = Hom RngCat U
23 20 21 2 22 rngchomfval φ Hom RngCat U = RngHomo Base RngCat U × Base RngCat U
24 20 21 2 rngcbas φ Base RngCat U = U Rng
25 incom Rng U = U Rng
26 3 25 eqtrdi φ B = U Rng
27 26 eqcomd φ U Rng = B
28 24 27 eqtrd φ Base RngCat U = B
29 28 sqxpeqd φ Base RngCat U × Base RngCat U = B × B
30 29 reseq2d φ RngHomo Base RngCat U × Base RngCat U = RngHomo B × B
31 23 30 eqtrd φ Hom RngCat U = RngHomo B × B
32 31 adantr φ x B Hom RngCat U = RngHomo B × B
33 32 eqcomd φ x B RngHomo B × B = Hom RngCat U
34 33 oveqd φ x B x RngHomo B × B x = x Hom RngCat U x
35 26 eleq2d φ x B x U Rng
36 35 biimpa φ x B x U Rng
37 24 adantr φ x B Base RngCat U = U Rng
38 36 37 eleqtrrd φ x B x Base RngCat U
39 20 21 14 22 38 38 rngchom φ x B x Hom RngCat U x = x RngHomo x
40 19 34 39 3eqtrd φ x B x H x = x RngHomo x
41 12 18 40 3eltr4d φ x B Id C x x H x