Metamath Proof Explorer


Theorem ghmmhm

Description: A group homomorphism is a monoid homomorphism. (Contributed by Stefan O'Rear, 7-Mar-2015)

Ref Expression
Assertion ghmmhm ⊢ F ∈ S GrpHom T → F ∈ S MndHom T

Proof

Step Hyp Ref Expression
1 ghmgrp1 ⊢ F ∈ S GrpHom T → S ∈ Grp
2 1 grpmndd ⊢ F ∈ S GrpHom T → S ∈ Mnd
3 ghmgrp2 ⊢ F ∈ S GrpHom T → T ∈ Grp
4 3 grpmndd ⊢ F ∈ S GrpHom T → T ∈ Mnd
5 eqid ⊢ Base S = Base S
6 eqid ⊢ Base T = Base T
7 5 6 ghmf ⊢ F ∈ S GrpHom T → F : Base S ⟶ Base T
8 eqid ⊢ + S = + S
9 eqid ⊢ + T = + T
10 5 8 9 ghmlin ⊢ F ∈ S GrpHom T ∧ x ∈ Base S ∧ y ∈ Base S → F ⁡ x + S y = F ⁡ x + T F ⁡ y
11 10 3expb ⊢ F ∈ S GrpHom T ∧ x ∈ Base S ∧ y ∈ Base S → F ⁡ x + S y = F ⁡ x + T F ⁡ y
12 11 ralrimivva ⊢ F ∈ S GrpHom T → ∀ x ∈ Base S ∀ y ∈ Base S F ⁡ x + S y = F ⁡ x + T F ⁡ y
13 eqid ⊢ 0 S = 0 S
14 eqid ⊢ 0 T = 0 T
15 13 14 ghmid ⊢ F ∈ S GrpHom T → F ⁡ 0 S = 0 T
16 7 12 15 3jca ⊢ F ∈ S GrpHom T → F : Base S ⟶ Base T ∧ ∀ x ∈ Base S ∀ y ∈ Base S F ⁡ x + S y = F ⁡ x + T F ⁡ y ∧ F ⁡ 0 S = 0 T
17 5 6 8 9 13 14 ismhm ⊢ F ∈ S MndHom T ↔ S ∈ Mnd ∧ T ∈ Mnd ∧ F : Base S ⟶ Base T ∧ ∀ x ∈ Base S ∀ y ∈ Base S F ⁡ x + S y = F ⁡ x + T F ⁡ y ∧ F ⁡ 0 S = 0 T
18 2 4 16 17 syl21anbrc ⊢ F ∈ S GrpHom T → F ∈ S MndHom T