Metamath Proof Explorer


Theorem ply1moneq

Description: Two monomials are equal iff their powers are equal. (Contributed by Thierry Arnoux, 20-Feb-2025)

Ref Expression
Hypotheses ply1moneq.p P=Poly1R
ply1moneq.x X=var1R
ply1moneq.e ×˙=mulGrpP
ply1moneq.r φRNzRing
ply1moneq.m φM0
ply1moneq.n φN0
Assertion ply1moneq φM×˙X=N×˙XM=N

Proof

Step Hyp Ref Expression
1 ply1moneq.p P=Poly1R
2 ply1moneq.x X=var1R
3 ply1moneq.e ×˙=mulGrpP
4 ply1moneq.r φRNzRing
5 ply1moneq.m φM0
6 ply1moneq.n φN0
7 nzrring RNzRingRRing
8 4 7 syl φRRing
9 eqid 0R=0R
10 eqid 1R=1R
11 1 2 3 8 5 9 10 coe1mon φcoe1M×˙X=k0ifk=M1R0R
12 fvexd φk01RV
13 fvexd φk00RV
14 12 13 ifcld φk0ifk=M1R0RV
15 11 14 fvmpt2d φk0coe1M×˙Xk=ifk=M1R0R
16 1 2 3 8 6 9 10 coe1mon φcoe1N×˙X=k0ifk=N1R0R
17 12 13 ifcld φk0ifk=N1R0RV
18 16 17 fvmpt2d φk0coe1N×˙Xk=ifk=N1R0R
19 15 18 eqeq12d φk0coe1M×˙Xk=coe1N×˙Xkifk=M1R0R=ifk=N1R0R
20 10 9 nzrnz RNzRing1R0R
21 4 20 syl φ1R0R
22 21 adantr φk01R0R
23 ifnebib 1R0Rifk=M1R0R=ifk=N1R0Rk=Mk=N
24 22 23 syl φk0ifk=M1R0R=ifk=N1R0Rk=Mk=N
25 19 24 bitrd φk0coe1M×˙Xk=coe1N×˙Xkk=Mk=N
26 25 ralbidva φk0coe1M×˙Xk=coe1N×˙Xkk0k=Mk=N
27 eqid mulGrpP=mulGrpP
28 eqid BaseP=BaseP
29 1 2 27 3 28 ply1moncl RRingM0M×˙XBaseP
30 8 5 29 syl2anc φM×˙XBaseP
31 1 2 27 3 28 ply1moncl RRingN0N×˙XBaseP
32 8 6 31 syl2anc φN×˙XBaseP
33 eqid coe1M×˙X=coe1M×˙X
34 eqid coe1N×˙X=coe1N×˙X
35 1 28 33 34 ply1coe1eq RRingM×˙XBasePN×˙XBasePk0coe1M×˙Xk=coe1N×˙XkM×˙X=N×˙X
36 8 30 32 35 syl3anc φk0coe1M×˙Xk=coe1N×˙XkM×˙X=N×˙X
37 5 6 eqelbid φk0k=Mk=NM=N
38 26 36 37 3bitr3d φM×˙X=N×˙XM=N