Metamath Proof Explorer


Theorem rhmsubclem1

Description: Lemma 1 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 rhmsubclem1 φ H Fn R × R

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 eqid x R , y R x GrpHom y mulGrp x MndHom mulGrp y = x R , y R x GrpHom y mulGrp x MndHom mulGrp y
6 ovex x GrpHom y V
7 6 inex1 x GrpHom y mulGrp x MndHom mulGrp y V
8 5 7 fnmpoi x R , y R x GrpHom y mulGrp x MndHom mulGrp y Fn R × R
9 4 a1i φ H = RingHom R × R
10 dfrhm2 RingHom = x Ring , y Ring x GrpHom y mulGrp x MndHom mulGrp y
11 10 a1i φ RingHom = x Ring , y Ring x GrpHom y mulGrp x MndHom mulGrp y
12 11 reseq1d φ RingHom R × R = x Ring , y Ring x GrpHom y mulGrp x MndHom mulGrp y R × R
13 inss1 Ring U Ring
14 3 13 eqsstrdi φ R Ring
15 resmpo R Ring R Ring x Ring , y Ring x GrpHom y mulGrp x MndHom mulGrp y R × R = x R , y R x GrpHom y mulGrp x MndHom mulGrp y
16 14 14 15 syl2anc φ x Ring , y Ring x GrpHom y mulGrp x MndHom mulGrp y R × R = x R , y R x GrpHom y mulGrp x MndHom mulGrp y
17 9 12 16 3eqtrd φ H = x R , y R x GrpHom y mulGrp x MndHom mulGrp y
18 17 fneq1d φ H Fn R × R x R , y R x GrpHom y mulGrp x MndHom mulGrp y Fn R × R
19 8 18 mpbiri φ H Fn R × R