Metamath Proof Explorer


Theorem mhmcompl

Description: The composition of a monoid homomorphism and a polynomial is a polynomial. (Contributed by SN, 7-Feb-2025)

Ref Expression
Hypotheses mhmcompl.p P = I mPoly R
mhmcompl.q Q = I mPoly S
mhmcompl.b B = Base P
mhmcompl.c C = Base Q
mhmcompl.h φ H R MndHom S
mhmcompl.f φ F B
Assertion mhmcompl φ H F C

Proof

Step Hyp Ref Expression
1 mhmcompl.p P = I mPoly R
2 mhmcompl.q Q = I mPoly S
3 mhmcompl.b B = Base P
4 mhmcompl.c C = Base Q
5 mhmcompl.h φ H R MndHom S
6 mhmcompl.f φ F B
7 fvexd φ Base S V
8 eqid f 0 I | f -1 Fin = f 0 I | f -1 Fin
9 ovexd φ 0 I V
10 8 9 rabexd φ f 0 I | f -1 Fin V
11 eqid Base R = Base R
12 eqid Base S = Base S
13 11 12 mhmf H R MndHom S H : Base R Base S
14 5 13 syl φ H : Base R Base S
15 1 11 3 8 6 mplelf φ F : f 0 I | f -1 Fin Base R
16 14 15 fcod φ H F : f 0 I | f -1 Fin Base S
17 7 10 16 elmapdd φ H F Base S f 0 I | f -1 Fin
18 eqid I mPwSer S = I mPwSer S
19 eqid Base I mPwSer S = Base I mPwSer S
20 1 3 mplrcl F B I V
21 6 20 syl φ I V
22 18 12 8 19 21 psrbas φ Base I mPwSer S = Base S f 0 I | f -1 Fin
23 17 22 eleqtrrd φ H F Base I mPwSer S
24 fvexd φ 0 S V
25 mhmrcl1 H R MndHom S R Mnd
26 5 25 syl φ R Mnd
27 eqid 0 R = 0 R
28 11 27 mndidcl R Mnd 0 R Base R
29 26 28 syl φ 0 R Base R
30 ssidd φ Base R Base R
31 fvexd φ Base R V
32 1 3 27 6 26 mplelsfi φ finSupp 0 R F
33 eqid 0 S = 0 S
34 27 33 mhm0 H R MndHom S H 0 R = 0 S
35 5 34 syl φ H 0 R = 0 S
36 24 29 15 14 30 10 31 32 35 fsuppcor φ finSupp 0 S H F
37 2 18 19 33 4 mplelbas H F C H F Base I mPwSer S finSupp 0 S H F
38 23 36 37 sylanbrc φ H F C