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=RevalSub1S
irngval.u U=R𝑠S
irngval.b B=BaseR
irngval.0 0˙=0R
elirng.r φRCRing
elirng.s φSSubRingR
0ringirng.1 φ¬RNzRing
Assertion 0ringirng φRIntgRingS=

Proof

Step Hyp Ref Expression
1 irngval.o O=RevalSub1S
2 irngval.u U=R𝑠S
3 irngval.b B=BaseR
4 irngval.0 0˙=0R
5 elirng.r φRCRing
6 elirng.s φSSubRingR
7 0ringirng.1 φ¬RNzRing
8 rex0 ¬pOpx=0˙
9 eqid Monic1pU=Monic1pU
10 eqid BaseU=BaseU
11 2 subrgring SSubRingRURing
12 6 11 syl φURing
13 5 crngringd φRRing
14 3 fveq2i B=BaseR
15 0ringnnzr RRingBaseR=1¬RNzRing
16 15 biimpar RRing¬RNzRingBaseR=1
17 13 7 16 syl2anc φBaseR=1
18 14 17 eqtrid φB=1
19 3 subrgss SSubRingRSB
20 2 3 ressbas2 SBS=BaseU
21 6 19 20 3syl φS=BaseU
22 21 6 eqeltrrd φBaseUSubRingR
23 3 13 18 22 0ringsubrg φBaseU=1
24 9 10 12 23 0ringmon1p φMonic1pU=
25 24 rexeqdv φpMonic1pUOpx=0˙pOpx=0˙
26 8 25 mtbiri φ¬pMonic1pUOpx=0˙
27 1 2 3 4 5 6 elirng φxRIntgRingSxBpMonic1pUOpx=0˙
28 27 simplbda φxRIntgRingSpMonic1pUOpx=0˙
29 26 28 mtand φ¬xRIntgRingS
30 29 eq0rdv φRIntgRingS=