Metamath Proof Explorer


Theorem qus0subgbas

Description: The base set of a quotient of a group by the trivial (zero) subgroup. (Contributed by AV, 26-Feb-2025)

Ref Expression
Hypotheses qus0subg.0 ⊢ 0 ˙ = 0 G
qus0subg.s ⊢ S = 0 ˙
qus0subg.e ⊢ ∼ ˙ = G ~ QG S
qus0subg.u ⊢ U = G / 𝑠 ∼ ˙
qus0subg.b ⊢ B = Base G
Assertion qus0subgbas ⊢ G ∈ Grp → Base U = u | ∃ x ∈ B u = x

Proof

Step Hyp Ref Expression
1 qus0subg.0 ⊢ 0 ˙ = 0 G
2 qus0subg.s ⊢ S = 0 ˙
3 qus0subg.e ⊢ ∼ ˙ = G ~ QG S
4 qus0subg.u ⊢ U = G / 𝑠 ∼ ˙
5 qus0subg.b ⊢ B = Base G
6 df-qs ⊢ B / ∼ ˙ = u | ∃ x ∈ B u = x ∼ ˙
7 4 a1i ⊢ G ∈ Grp → U = G / 𝑠 ∼ ˙
8 5 a1i ⊢ G ∈ Grp → B = Base G
9 3 ovexi ⊢ ∼ ˙ ∈ V
10 9 a1i ⊢ G ∈ Grp → ∼ ˙ ∈ V
11 id ⊢ G ∈ Grp → G ∈ Grp
12 7 8 10 11 qusbas ⊢ G ∈ Grp → B / ∼ ˙ = Base U
13 1 2 5 3 eqg0subgecsn ⊢ G ∈ Grp ∧ x ∈ B → x ∼ ˙ = x
14 13 eqeq2d ⊢ G ∈ Grp ∧ x ∈ B → u = x ∼ ˙ ↔ u = x
15 14 rexbidva ⊢ G ∈ Grp → ∃ x ∈ B u = x ∼ ˙ ↔ ∃ x ∈ B u = x
16 15 abbidv ⊢ G ∈ Grp → u | ∃ x ∈ B u = x ∼ ˙ = u | ∃ x ∈ B u = x
17 6 12 16 3eqtr3a ⊢ G ∈ Grp → Base U = u | ∃ x ∈ B u = x