Metamath Proof Explorer


Definition df-mhm

Description: A monoid homomorphism is a function on the base sets which preserves the binary operation and the identity. (Contributed by Mario Carneiro, 7-Mar-2015)

Ref Expression
Assertion df-mhm MndHom = s Mnd , t Mnd f Base t Base s | x Base s y Base s f x + s y = f x + t f y f 0 s = 0 t

Detailed syntax breakdown

Step Hyp Ref Expression
0 cmhm class MndHom
1 vs setvar s
2 cmnd class Mnd
3 vt setvar t
4 vf setvar f
5 cbs class Base
6 3 cv setvar t
7 6 5 cfv class Base t
8 cmap class 𝑚
9 1 cv setvar s
10 9 5 cfv class Base s
11 7 10 8 co class Base t Base s
12 vx setvar x
13 vy setvar y
14 4 cv setvar f
15 12 cv setvar x
16 cplusg class + 𝑔
17 9 16 cfv class + s
18 13 cv setvar y
19 15 18 17 co class x + s y
20 19 14 cfv class f x + s y
21 15 14 cfv class f x
22 6 16 cfv class + t
23 18 14 cfv class f y
24 21 23 22 co class f x + t f y
25 20 24 wceq wff f x + s y = f x + t f y
26 25 13 10 wral wff y Base s f x + s y = f x + t f y
27 26 12 10 wral wff x Base s y Base s f x + s y = f x + t f y
28 c0g class 0 𝑔
29 9 28 cfv class 0 s
30 29 14 cfv class f 0 s
31 6 28 cfv class 0 t
32 30 31 wceq wff f 0 s = 0 t
33 27 32 wa wff x Base s y Base s f x + s y = f x + t f y f 0 s = 0 t
34 33 4 11 crab class f Base t Base s | x Base s y Base s f x + s y = f x + t f y f 0 s = 0 t
35 1 3 2 2 34 cmpo class s Mnd , t Mnd f Base t Base s | x Base s y Base s f x + s y = f x + t f y f 0 s = 0 t
36 0 35 wceq wff MndHom = s Mnd , t Mnd f Base t Base s | x Base s y Base s f x + s y = f x + t f y f 0 s = 0 t