Metamath Proof Explorer


Theorem isgrpd

Description: Deduce a group from its properties. Unlike isgrpd2 , this one goes straight from the base properties rather than going through Mnd . N (negative) is normally dependent on x i.e. read it as N ( x ) . (Contributed by NM, 6-Jun-2013) (Revised by Mario Carneiro, 6-Jan-2015)

Ref Expression
Hypotheses isgrpd.b ⊢ φ → B = Base G
isgrpd.p ⊢ φ → + ˙ = + G
isgrpd.c ⊢ φ ∧ x ∈ B ∧ y ∈ B → x + ˙ y ∈ B
isgrpd.a ⊢ φ ∧ x ∈ B ∧ y ∈ B ∧ z ∈ B → x + ˙ y + ˙ z = x + ˙ y + ˙ z
isgrpd.z ⊢ φ → 0 ˙ ∈ B
isgrpd.i ⊢ φ ∧ x ∈ B → 0 ˙ + ˙ x = x
isgrpd.n ⊢ φ ∧ x ∈ B → N ∈ B
isgrpd.j ⊢ φ ∧ x ∈ B → N + ˙ x = 0 ˙
Assertion isgrpd ⊢ φ → G ∈ Grp

Proof

Step Hyp Ref Expression
1 isgrpd.b ⊢ φ → B = Base G
2 isgrpd.p ⊢ φ → + ˙ = + G
3 isgrpd.c ⊢ φ ∧ x ∈ B ∧ y ∈ B → x + ˙ y ∈ B
4 isgrpd.a ⊢ φ ∧ x ∈ B ∧ y ∈ B ∧ z ∈ B → x + ˙ y + ˙ z = x + ˙ y + ˙ z
5 isgrpd.z ⊢ φ → 0 ˙ ∈ B
6 isgrpd.i ⊢ φ ∧ x ∈ B → 0 ˙ + ˙ x = x
7 isgrpd.n ⊢ φ ∧ x ∈ B → N ∈ B
8 isgrpd.j ⊢ φ ∧ x ∈ B → N + ˙ x = 0 ˙
9 oveq1 ⊢ y = N → y + ˙ x = N + ˙ x
10 9 eqeq1d ⊢ y = N → y + ˙ x = 0 ˙ ↔ N + ˙ x = 0 ˙
11 10 rspcev ⊢ N ∈ B ∧ N + ˙ x = 0 ˙ → ∃ y ∈ B y + ˙ x = 0 ˙
12 7 8 11 syl2anc ⊢ φ ∧ x ∈ B → ∃ y ∈ B y + ˙ x = 0 ˙
13 1 2 3 4 5 6 12 isgrpde ⊢ φ → G ∈ Grp