Metamath Proof Explorer


Theorem rhmdvdsr

Description: A ring homomorphism preserves the divisibility relation. (Contributed by Thierry Arnoux, 22-Oct-2017)

Ref Expression
Hypotheses rhmdvdsr.x X=BaseR
rhmdvdsr.m ˙=rR
rhmdvdsr.n ×˙=rS
Assertion rhmdvdsr FRRingHomSAXBXA˙BFA×˙FB

Proof

Step Hyp Ref Expression
1 rhmdvdsr.x X=BaseR
2 rhmdvdsr.m ˙=rR
3 rhmdvdsr.n ×˙=rS
4 simpl1 FRRingHomSAXBXA˙BFRRingHomS
5 simpl2 FRRingHomSAXBXA˙BAX
6 eqid BaseS=BaseS
7 1 6 rhmf FRRingHomSF:XBaseS
8 7 ffvelcdmda FRRingHomSAXFABaseS
9 4 5 8 syl2anc FRRingHomSAXBXA˙BFABaseS
10 simpll1 FRRingHomSAXBXA˙BcXFRRingHomS
11 simpr FRRingHomSAXBXA˙BcXcX
12 7 ffvelcdmda FRRingHomScXFcBaseS
13 10 11 12 syl2anc FRRingHomSAXBXA˙BcXFcBaseS
14 13 ralrimiva FRRingHomSAXBXA˙BcXFcBaseS
15 5 adantr FRRingHomSAXBXA˙BcXAX
16 eqid R=R
17 eqid S=S
18 1 16 17 rhmmul FRRingHomScXAXFcRA=FcSFA
19 10 11 15 18 syl3anc FRRingHomSAXBXA˙BcXFcRA=FcSFA
20 19 ralrimiva FRRingHomSAXBXA˙BcXFcRA=FcSFA
21 simpr FRRingHomSAXBXA˙BA˙B
22 1 2 16 dvdsr2 AXA˙BcXcRA=B
23 22 biimpac A˙BAXcXcRA=B
24 21 5 23 syl2anc FRRingHomSAXBXA˙BcXcRA=B
25 r19.29 cXFcRA=FcSFAcXcRA=BcXFcRA=FcSFAcRA=B
26 simpl FcRA=FcSFAcRA=BFcRA=FcSFA
27 simpr FcRA=FcSFAcRA=BcRA=B
28 27 fveq2d FcRA=FcSFAcRA=BFcRA=FB
29 26 28 eqtr3d FcRA=FcSFAcRA=BFcSFA=FB
30 29 reximi cXFcRA=FcSFAcRA=BcXFcSFA=FB
31 25 30 syl cXFcRA=FcSFAcXcRA=BcXFcSFA=FB
32 20 24 31 syl2anc FRRingHomSAXBXA˙BcXFcSFA=FB
33 r19.29 cXFcBaseScXFcSFA=FBcXFcBaseSFcSFA=FB
34 14 32 33 syl2anc FRRingHomSAXBXA˙BcXFcBaseSFcSFA=FB
35 oveq1 y=FcySFA=FcSFA
36 35 eqeq1d y=FcySFA=FBFcSFA=FB
37 36 rspcev FcBaseSFcSFA=FByBaseSySFA=FB
38 37 rexlimivw cXFcBaseSFcSFA=FByBaseSySFA=FB
39 34 38 syl FRRingHomSAXBXA˙ByBaseSySFA=FB
40 6 3 17 dvdsr FA×˙FBFABaseSyBaseSySFA=FB
41 9 39 40 sylanbrc FRRingHomSAXBXA˙BFA×˙FB