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 = Poly 1 R
ply1moneq.x X = var 1 R
ply1moneq.e × ˙ = mulGrp P
ply1moneq.r φ R NzRing
ply1moneq.m φ M 0
ply1moneq.n φ N 0
Assertion ply1moneq φ M × ˙ X = N × ˙ X M = N

Proof

Step Hyp Ref Expression
1 ply1moneq.p P = Poly 1 R
2 ply1moneq.x X = var 1 R
3 ply1moneq.e × ˙ = mulGrp P
4 ply1moneq.r φ R NzRing
5 ply1moneq.m φ M 0
6 ply1moneq.n φ N 0
7 nzrring R NzRing R Ring
8 4 7 syl φ R Ring
9 eqid 0 R = 0 R
10 eqid 1 R = 1 R
11 1 2 3 8 5 9 10 coe1mon φ coe 1 M × ˙ X = k 0 if k = M 1 R 0 R
12 fvexd φ k 0 1 R V
13 fvexd φ k 0 0 R V
14 12 13 ifcld φ k 0 if k = M 1 R 0 R V
15 11 14 fvmpt2d φ k 0 coe 1 M × ˙ X k = if k = M 1 R 0 R
16 1 2 3 8 6 9 10 coe1mon φ coe 1 N × ˙ X = k 0 if k = N 1 R 0 R
17 12 13 ifcld φ k 0 if k = N 1 R 0 R V
18 16 17 fvmpt2d φ k 0 coe 1 N × ˙ X k = if k = N 1 R 0 R
19 15 18 eqeq12d φ k 0 coe 1 M × ˙ X k = coe 1 N × ˙ X k if k = M 1 R 0 R = if k = N 1 R 0 R
20 10 9 nzrnz R NzRing 1 R 0 R
21 4 20 syl φ 1 R 0 R
22 21 adantr φ k 0 1 R 0 R
23 ifnebib 1 R 0 R if k = M 1 R 0 R = if k = N 1 R 0 R k = M k = N
24 22 23 syl φ k 0 if k = M 1 R 0 R = if k = N 1 R 0 R k = M k = N
25 19 24 bitrd φ k 0 coe 1 M × ˙ X k = coe 1 N × ˙ X k k = M k = N
26 25 ralbidva φ k 0 coe 1 M × ˙ X k = coe 1 N × ˙ X k k 0 k = M k = N
27 eqid mulGrp P = mulGrp P
28 eqid Base P = Base P
29 1 2 27 3 28 ply1moncl R Ring M 0 M × ˙ X Base P
30 8 5 29 syl2anc φ M × ˙ X Base P
31 1 2 27 3 28 ply1moncl R Ring N 0 N × ˙ X Base P
32 8 6 31 syl2anc φ N × ˙ X Base P
33 eqid coe 1 M × ˙ X = coe 1 M × ˙ X
34 eqid coe 1 N × ˙ X = coe 1 N × ˙ X
35 1 28 33 34 ply1coe1eq R Ring M × ˙ X Base P N × ˙ X Base P k 0 coe 1 M × ˙ X k = coe 1 N × ˙ X k M × ˙ X = N × ˙ X
36 8 30 32 35 syl3anc φ k 0 coe 1 M × ˙ X k = coe 1 N × ˙ X k M × ˙ X = N × ˙ X
37 5 6 eqelbid φ k 0 k = M k = N M = N
38 26 36 37 3bitr3d φ M × ˙ X = N × ˙ X M = N