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Metamath Proof Explorer

Theorem lsmelvalx

Description: Subspace sum membership (for a group or vector space). Extended domain version of lsmelval . (Contributed by NM, 28-Jan-2014) (Revised by Mario Carneiro, 19-Apr-2016)

Ref Expression
Hypotheses lsmfval.v ⊢ 𝐵 = ( Base ‘ 𝐺 )
lsmfval.a ⊢ + = ( +g ‘ 𝐺 )
lsmfval.s ⊢ ⊕ = ( LSSum ‘ 𝐺 )
Assertion lsmelvalx ( ( 𝐺 ∈ 𝑉 ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵 ) → ( 𝑋 ∈ ( 𝑇 ⊕ 𝑈 ) ↔ ∃ 𝑦 ∈ 𝑇 ∃ 𝑧 ∈ 𝑈 𝑋 = ( 𝑦 + 𝑧 ) ) )

Proof

Step Hyp Ref Expression
1 lsmfval.v ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 lsmfval.a ⊢ + = ( +g ‘ 𝐺 )
3 lsmfval.s ⊢ ⊕ = ( LSSum ‘ 𝐺 )
4 1 2 3 lsmvalx ⊢ ( ( 𝐺 ∈ 𝑉 ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵 ) → ( 𝑇 ⊕ 𝑈 ) = ran ( 𝑦 ∈ 𝑇 , 𝑧 ∈ 𝑈 ↦ ( 𝑦 + 𝑧 ) ) )
5 4 eleq2d ⊢ ( ( 𝐺 ∈ 𝑉 ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵 ) → ( 𝑋 ∈ ( 𝑇 ⊕ 𝑈 ) ↔ 𝑋 ∈ ran ( 𝑦 ∈ 𝑇 , 𝑧 ∈ 𝑈 ↦ ( 𝑦 + 𝑧 ) ) ) )
6 eqid ⊢ ( 𝑦 ∈ 𝑇 , 𝑧 ∈ 𝑈 ↦ ( 𝑦 + 𝑧 ) ) = ( 𝑦 ∈ 𝑇 , 𝑧 ∈ 𝑈 ↦ ( 𝑦 + 𝑧 ) )
7 ovex ⊢ ( 𝑦 + 𝑧 ) ∈ V
8 6 7 elrnmpo ⊢ ( 𝑋 ∈ ran ( 𝑦 ∈ 𝑇 , 𝑧 ∈ 𝑈 ↦ ( 𝑦 + 𝑧 ) ) ↔ ∃ 𝑦 ∈ 𝑇 ∃ 𝑧 ∈ 𝑈 𝑋 = ( 𝑦 + 𝑧 ) )
9 5 8 bitrdi ⊢ ( ( 𝐺 ∈ 𝑉 ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵 ) → ( 𝑋 ∈ ( 𝑇 ⊕ 𝑈 ) ↔ ∃ 𝑦 ∈ 𝑇 ∃ 𝑧 ∈ 𝑈 𝑋 = ( 𝑦 + 𝑧 ) ) )