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Metamath Proof Explorer

Theorem dpjlsm

Description: The two subgroups that appear in dpjval add to the full direct product. (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 )
dpjlsm.s ⊢ ⊕ = ( LSSum ‘ 𝐺 )
Assertion dpjlsm ( 𝜑 → ( 𝐺 DProd 𝑆 ) = ( ( 𝑆 ‘ 𝑋 ) ⊕ ( 𝐺 DProd ( 𝑆 ↾ ( 𝐼 ∖ { 𝑋 } ) ) ) ) )

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 dpjlsm.s ⊢ ⊕ = ( LSSum ‘ 𝐺 )
5 1 2 dprdf2 ⊢ ( 𝜑 → 𝑆 : 𝐼 ⟶ ( SubGrp ‘ 𝐺 ) )
6 disjdif ⊢ ( { 𝑋 } ∩ ( 𝐼 ∖ { 𝑋 } ) ) = ∅
7 6 a1i ⊢ ( 𝜑 → ( { 𝑋 } ∩ ( 𝐼 ∖ { 𝑋 } ) ) = ∅ )
8 undif2 ⊢ ( { 𝑋 } ∪ ( 𝐼 ∖ { 𝑋 } ) ) = ( { 𝑋 } ∪ 𝐼 )
9 3 snssd ⊢ ( 𝜑 → { 𝑋 } ⊆ 𝐼 )
10 ssequn1 ⊢ ( { 𝑋 } ⊆ 𝐼 ↔ ( { 𝑋 } ∪ 𝐼 ) = 𝐼 )
11 9 10 sylib ⊢ ( 𝜑 → ( { 𝑋 } ∪ 𝐼 ) = 𝐼 )
12 8 11 eqtr2id ⊢ ( 𝜑 → 𝐼 = ( { 𝑋 } ∪ ( 𝐼 ∖ { 𝑋 } ) ) )
13 5 7 12 4 1 dprdsplit ⊢ ( 𝜑 → ( 𝐺 DProd 𝑆 ) = ( ( 𝐺 DProd ( 𝑆 ↾ { 𝑋 } ) ) ⊕ ( 𝐺 DProd ( 𝑆 ↾ ( 𝐼 ∖ { 𝑋 } ) ) ) ) )
14 1 2 3 dpjlem ⊢ ( 𝜑 → ( 𝐺 DProd ( 𝑆 ↾ { 𝑋 } ) ) = ( 𝑆 ‘ 𝑋 ) )
15 14 oveq1d ⊢ ( 𝜑 → ( ( 𝐺 DProd ( 𝑆 ↾ { 𝑋 } ) ) ⊕ ( 𝐺 DProd ( 𝑆 ↾ ( 𝐼 ∖ { 𝑋 } ) ) ) ) = ( ( 𝑆 ‘ 𝑋 ) ⊕ ( 𝐺 DProd ( 𝑆 ↾ ( 𝐼 ∖ { 𝑋 } ) ) ) ) )
16 13 15 eqtrd ⊢ ( 𝜑 → ( 𝐺 DProd 𝑆 ) = ( ( 𝑆 ‘ 𝑋 ) ⊕ ( 𝐺 DProd ( 𝑆 ↾ ( 𝐼 ∖ { 𝑋 } ) ) ) ) )