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Metamath Proof Explorer

Theorem fsumsub

Description: Split a finite sum over a subtraction. (Contributed by Scott Fenton, 12-Jun-2013) (Revised by Mario Carneiro, 24-Apr-2014)

Ref Expression
Hypotheses fsumneg.1 ⊢ ( 𝜑 → 𝐴 ∈ Fin )
fsumneg.2 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
fsumsub.3 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐶 ∈ ℂ )
Assertion fsumsub ( 𝜑 → Σ 𝑘 ∈ 𝐴 ( 𝐵 − 𝐶 ) = ( Σ 𝑘 ∈ 𝐴 𝐵 − Σ 𝑘 ∈ 𝐴 𝐶 ) )

Proof

Step Hyp Ref Expression
1 fsumneg.1 ⊢ ( 𝜑 → 𝐴 ∈ Fin )
2 fsumneg.2 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
3 fsumsub.3 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐶 ∈ ℂ )
4 3 negcld ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → - 𝐶 ∈ ℂ )
5 1 2 4 fsumadd ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 ( 𝐵 + - 𝐶 ) = ( Σ 𝑘 ∈ 𝐴 𝐵 + Σ 𝑘 ∈ 𝐴 - 𝐶 ) )
6 1 3 fsumneg ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 - 𝐶 = - Σ 𝑘 ∈ 𝐴 𝐶 )
7 6 oveq2d ⊢ ( 𝜑 → ( Σ 𝑘 ∈ 𝐴 𝐵 + Σ 𝑘 ∈ 𝐴 - 𝐶 ) = ( Σ 𝑘 ∈ 𝐴 𝐵 + - Σ 𝑘 ∈ 𝐴 𝐶 ) )
8 5 7 eqtrd ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 ( 𝐵 + - 𝐶 ) = ( Σ 𝑘 ∈ 𝐴 𝐵 + - Σ 𝑘 ∈ 𝐴 𝐶 ) )
9 2 3 negsubd ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → ( 𝐵 + - 𝐶 ) = ( 𝐵 − 𝐶 ) )
10 9 sumeq2dv ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 ( 𝐵 + - 𝐶 ) = Σ 𝑘 ∈ 𝐴 ( 𝐵 − 𝐶 ) )
11 1 2 fsumcl ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 𝐵 ∈ ℂ )
12 1 3 fsumcl ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 𝐶 ∈ ℂ )
13 11 12 negsubd ⊢ ( 𝜑 → ( Σ 𝑘 ∈ 𝐴 𝐵 + - Σ 𝑘 ∈ 𝐴 𝐶 ) = ( Σ 𝑘 ∈ 𝐴 𝐵 − Σ 𝑘 ∈ 𝐴 𝐶 ) )
14 8 10 13 3eqtr3d ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 ( 𝐵 − 𝐶 ) = ( Σ 𝑘 ∈ 𝐴 𝐵 − Σ 𝑘 ∈ 𝐴 𝐶 ) )