Metamath Proof Explorer


Theorem ghmfghm

Description: The function fulfilling the conditions of ghmgrp is a group homomorphism. (Contributed by Thierry Arnoux, 26-Jan-2020)

Ref Expression
Hypotheses ghmabl.x ⊢ X = Base G
ghmabl.y ⊢ Y = Base H
ghmabl.p ⊢ + ˙ = + G
ghmabl.q ⊢ ⨣ ˙ = + H
ghmabl.f ⊢ φ ∧ x ∈ X ∧ y ∈ X → F ⁡ x + ˙ y = F ⁡ x ⨣ ˙ F ⁡ y
ghmabl.1 ⊢ φ → F : X ⟶ onto Y
ghmfghm.3 ⊢ φ → G ∈ Grp
Assertion ghmfghm ⊢ φ → F ∈ G GrpHom H

Proof

Step Hyp Ref Expression
1 ghmabl.x ⊢ X = Base G
2 ghmabl.y ⊢ Y = Base H
3 ghmabl.p ⊢ + ˙ = + G
4 ghmabl.q ⊢ ⨣ ˙ = + H
5 ghmabl.f ⊢ φ ∧ x ∈ X ∧ y ∈ X → F ⁡ x + ˙ y = F ⁡ x ⨣ ˙ F ⁡ y
6 ghmabl.1 ⊢ φ → F : X ⟶ onto Y
7 ghmfghm.3 ⊢ φ → G ∈ Grp
8 5 1 2 3 4 6 7 ghmgrp ⊢ φ → H ∈ Grp
9 fof ⊢ F : X ⟶ onto Y → F : X ⟶ Y
10 6 9 syl ⊢ φ → F : X ⟶ Y
11 5 3expb ⊢ φ ∧ x ∈ X ∧ y ∈ X → F ⁡ x + ˙ y = F ⁡ x ⨣ ˙ F ⁡ y
12 1 2 3 4 7 8 10 11 isghmd ⊢ φ → F ∈ G GrpHom H