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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 = ( 0g ‘ 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 = ( 0g ‘ 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 } )