Metamath Proof Explorer


Theorem 0ringirng

Description: A zero ring R has no integral elements. (Contributed by Thierry Arnoux, 5-Feb-2025)

Ref Expression
Hypotheses irngval.o O = R evalSub 1 S
irngval.u U = R 𝑠 S
irngval.b B = Base R
irngval.0 0 ˙ = 0 R
elirng.r φ R CRing
elirng.s φ S SubRing R
0ringirng.1 φ ¬ R NzRing
Assertion 0ringirng φ R IntgRing S =

Proof

Step Hyp Ref Expression
1 irngval.o O = R evalSub 1 S
2 irngval.u U = R 𝑠 S
3 irngval.b B = Base R
4 irngval.0 0 ˙ = 0 R
5 elirng.r φ R CRing
6 elirng.s φ S SubRing R
7 0ringirng.1 φ ¬ R NzRing
8 rex0 ¬ p O p x = 0 ˙
9 eqid Monic 1p U = Monic 1p U
10 eqid Base U = Base U
11 2 subrgring S SubRing R U Ring
12 6 11 syl φ U Ring
13 5 crngringd φ R Ring
14 3 fveq2i B = Base R
15 0ringnnzr R Ring Base R = 1 ¬ R NzRing
16 15 biimpar R Ring ¬ R NzRing Base R = 1
17 13 7 16 syl2anc φ Base R = 1
18 14 17 eqtrid φ B = 1
19 3 subrgss S SubRing R S B
20 2 3 ressbas2 S B S = Base U
21 6 19 20 3syl φ S = Base U
22 21 6 eqeltrrd φ Base U SubRing R
23 3 13 18 22 0ringsubrg φ Base U = 1
24 9 10 12 23 0ringmon1p φ Monic 1p U =
25 24 rexeqdv φ p Monic 1p U O p x = 0 ˙ p O p x = 0 ˙
26 8 25 mtbiri φ ¬ p Monic 1p U O p x = 0 ˙
27 1 2 3 4 5 6 elirng φ x R IntgRing S x B p Monic 1p U O p x = 0 ˙
28 27 simplbda φ x R IntgRing S p Monic 1p U O p x = 0 ˙
29 26 28 mtand φ ¬ x R IntgRing S
30 29 eq0rdv φ R IntgRing S =