Metamath Proof Explorer


Theorem dpjdisj

Description: The two subgroups that appear in dpjval are disjoint. (Contributed by Mario Carneiro, 26-Apr-2016)

Ref Expression
Hypotheses dpjfval.1 ⊢ φ → G dom ⁡ DProd S
dpjfval.2 ⊢ φ → dom ⁡ S = I
dpjlem.3 ⊢ φ → X ∈ I
dpjdisj.0 ⊢ 0 ˙ = 0 G
Assertion dpjdisj ⊢ φ → S ⁡ X ∩ G DProd S ↾ I ∖ X = 0 ˙

Proof

Step Hyp Ref Expression
1 dpjfval.1 ⊢ φ → G dom ⁡ DProd S
2 dpjfval.2 ⊢ φ → dom ⁡ S = I
3 dpjlem.3 ⊢ φ → X ∈ I
4 dpjdisj.0 ⊢ 0 ˙ = 0 G
5 1 2 3 dpjlem ⊢ φ → G DProd S ↾ X = S ⁡ X
6 5 ineq1d ⊢ φ → G DProd S ↾ X ∩ G DProd S ↾ I ∖ X = S ⁡ X ∩ G DProd S ↾ I ∖ X
7 1 2 dprdf2 ⊢ φ → S : I ⟶ SubGrp ⁡ G
8 disjdif ⊢ X ∩ I ∖ X = ∅
9 8 a1i ⊢ φ → X ∩ I ∖ X = ∅
10 undif2 ⊢ X ∪ I ∖ X = X ∪ I
11 3 snssd ⊢ φ → X ⊆ I
12 ssequn1 ⊢ X ⊆ I ↔ X ∪ I = I
13 11 12 sylib ⊢ φ → X ∪ I = I
14 10 13 eqtr2id ⊢ φ → I = X ∪ I ∖ X
15 eqid ⊢ Cntz ⁡ G = Cntz ⁡ G
16 7 9 14 15 4 dmdprdsplit ⊢ φ → G dom ⁡ DProd S ↔ G dom ⁡ DProd S ↾ X ∧ G dom ⁡ DProd S ↾ I ∖ X ∧ G DProd S ↾ X ⊆ Cntz ⁡ G ⁡ G DProd S ↾ I ∖ X ∧ G DProd S ↾ X ∩ G DProd S ↾ I ∖ X = 0 ˙
17 1 16 mpbid ⊢ φ → G dom ⁡ DProd S ↾ X ∧ G dom ⁡ DProd S ↾ I ∖ X ∧ G DProd S ↾ X ⊆ Cntz ⁡ G ⁡ G DProd S ↾ I ∖ X ∧ G DProd S ↾ X ∩ G DProd S ↾ I ∖ X = 0 ˙
18 17 simp3d ⊢ φ → G DProd S ↾ X ∩ G DProd S ↾ I ∖ X = 0 ˙
19 6 18 eqtr3d ⊢ φ → S ⁡ X ∩ G DProd S ↾ I ∖ X = 0 ˙