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 ( 𝜑𝐴 ∈ Fin )
fsumneg.2 ( ( 𝜑𝑘𝐴 ) → 𝐵 ∈ ℂ )
fsumsub.3 ( ( 𝜑𝑘𝐴 ) → 𝐶 ∈ ℂ )
Assertion fsumsub ( 𝜑 → Σ 𝑘𝐴 ( 𝐵𝐶 ) = ( Σ 𝑘𝐴 𝐵 − Σ 𝑘𝐴 𝐶 ) )

Proof

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