Metamath Proof Explorer


Theorem 0ringmon1p

Description: There are no monic polynomials over a zero ring. (Contributed by Thierry Arnoux, 5-Feb-2025)

Ref Expression
Hypotheses 0ringmon1p.1 M=Monic1pR
0ringmon1p.2 B=BaseR
0ringmon1p.3 φRRing
0ringmon1p.4 φB=1
Assertion 0ringmon1p φM=

Proof

Step Hyp Ref Expression
1 0ringmon1p.1 M=Monic1pR
2 0ringmon1p.2 B=BaseR
3 0ringmon1p.3 φRRing
4 0ringmon1p.4 φB=1
5 eqid Poly1R=Poly1R
6 eqid BasePoly1R=BasePoly1R
7 eqid 0Poly1R=0Poly1R
8 eqid deg1R=deg1R
9 eqid 1R=1R
10 5 6 7 8 1 9 ismon1p pMpBasePoly1Rp0Poly1Rcoe1pdeg1Rp=1R
11 10 biimpi pMpBasePoly1Rp0Poly1Rcoe1pdeg1Rp=1R
12 11 adantl φpMpBasePoly1Rp0Poly1Rcoe1pdeg1Rp=1R
13 12 simp3d φpMcoe1pdeg1Rp=1R
14 3 adantr φpMRRing
15 12 simp1d φpMpBasePoly1R
16 12 simp2d φpMp0Poly1R
17 eqid 0R=0R
18 eqid coe1p=coe1p
19 8 5 7 6 17 18 deg1ldg RRingpBasePoly1Rp0Poly1Rcoe1pdeg1Rp0R
20 14 15 16 19 syl3anc φpMcoe1pdeg1Rp0R
21 2 17 9 0ring01eq RRingB=10R=1R
22 3 4 21 syl2anc φ0R=1R
23 22 adantr φpM0R=1R
24 20 23 neeqtrd φpMcoe1pdeg1Rp1R
25 24 neneqd φpM¬coe1pdeg1Rp=1R
26 13 25 pm2.65da φ¬pM
27 26 eq0rdv φM=