Metamath Proof Explorer


Theorem srhmsubcALTVlem2

Description: Lemma 2 for srhmsubcALTV . (Contributed by AV, 19-Feb-2020) (New usage is discouraged.)

Ref Expression
Hypotheses srhmsubcALTV.s r S r Ring
srhmsubcALTV.c C = U S
srhmsubcALTV.j J = r C , s C r RingHom s
Assertion srhmsubcALTVlem2 U V X C Y C X J Y = X Hom RingCatALTV U Y

Proof

Step Hyp Ref Expression
1 srhmsubcALTV.s r S r Ring
2 srhmsubcALTV.c C = U S
3 srhmsubcALTV.j J = r C , s C r RingHom s
4 3 a1i U V X C Y C J = r C , s C r RingHom s
5 oveq12 r = X s = Y r RingHom s = X RingHom Y
6 5 adantl U V X C Y C r = X s = Y r RingHom s = X RingHom Y
7 simpl X C Y C X C
8 7 adantl U V X C Y C X C
9 simpr X C Y C Y C
10 9 adantl U V X C Y C Y C
11 ovexd U V X C Y C X RingHom Y V
12 4 6 8 10 11 ovmpod U V X C Y C X J Y = X RingHom Y
13 eqid RingCatALTV U = RingCatALTV U
14 eqid Base RingCatALTV U = Base RingCatALTV U
15 simpl U V X C Y C U V
16 eqid Hom RingCatALTV U = Hom RingCatALTV U
17 1 2 srhmsubcALTVlem1 U V X C X Base RingCatALTV U
18 7 17 sylan2 U V X C Y C X Base RingCatALTV U
19 1 2 srhmsubcALTVlem1 U V Y C Y Base RingCatALTV U
20 9 19 sylan2 U V X C Y C Y Base RingCatALTV U
21 13 14 15 16 18 20 ringchomALTV U V X C Y C X Hom RingCatALTV U Y = X RingHom Y
22 12 21 eqtr4d U V X C Y C X J Y = X Hom RingCatALTV U Y