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Metamath Proof Explorer

Theorem fsumdivc

Description: A finite sum divided by a constant. (Contributed by NM, 2-Jan-2006) (Revised by Mario Carneiro, 24-Apr-2014)

Ref Expression
Hypotheses fsummulc2.1 ⊢ ( 𝜑 → 𝐴 ∈ Fin )
fsummulc2.2 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
fsummulc2.3 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
fsumdivc.4 ⊢ ( 𝜑 → 𝐶 ≠ 0 )
Assertion fsumdivc ( 𝜑 → ( Σ 𝑘 ∈ 𝐴 𝐵 / 𝐶 ) = Σ 𝑘 ∈ 𝐴 ( 𝐵 / 𝐶 ) )

Proof

Step Hyp Ref Expression
1 fsummulc2.1 ⊢ ( 𝜑 → 𝐴 ∈ Fin )
2 fsummulc2.2 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
3 fsummulc2.3 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
4 fsumdivc.4 ⊢ ( 𝜑 → 𝐶 ≠ 0 )
5 2 4 reccld ⊢ ( 𝜑 → ( 1 / 𝐶 ) ∈ ℂ )
6 1 5 3 fsummulc1 ⊢ ( 𝜑 → ( Σ 𝑘 ∈ 𝐴 𝐵 · ( 1 / 𝐶 ) ) = Σ 𝑘 ∈ 𝐴 ( 𝐵 · ( 1 / 𝐶 ) ) )
7 1 3 fsumcl ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 𝐵 ∈ ℂ )
8 7 2 4 divrecd ⊢ ( 𝜑 → ( Σ 𝑘 ∈ 𝐴 𝐵 / 𝐶 ) = ( Σ 𝑘 ∈ 𝐴 𝐵 · ( 1 / 𝐶 ) ) )
9 2 adantr ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐶 ∈ ℂ )
10 4 adantr ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐶 ≠ 0 )
11 3 9 10 divrecd ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → ( 𝐵 / 𝐶 ) = ( 𝐵 · ( 1 / 𝐶 ) ) )
12 11 sumeq2dv ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 ( 𝐵 / 𝐶 ) = Σ 𝑘 ∈ 𝐴 ( 𝐵 · ( 1 / 𝐶 ) ) )
13 6 8 12 3eqtr4d ⊢ ( 𝜑 → ( Σ 𝑘 ∈ 𝐴 𝐵 / 𝐶 ) = Σ 𝑘 ∈ 𝐴 ( 𝐵 / 𝐶 ) )