Description: Negation of a finite sum. (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 ) | ||
| Assertion | fsumneg | ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 - 𝐵 = - Σ 𝑘 ∈ 𝐴 𝐵 ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | fsumneg.1 | ⊢ ( 𝜑 → 𝐴 ∈ Fin ) | |
| 2 | fsumneg.2 | ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ ℂ ) | |
| 3 | neg1cn | ⊢ - 1 ∈ ℂ | |
| 4 | 3 | a1i | ⊢ ( 𝜑 → - 1 ∈ ℂ ) |
| 5 | 1 4 2 | fsummulc2 | ⊢ ( 𝜑 → ( - 1 · Σ 𝑘 ∈ 𝐴 𝐵 ) = Σ 𝑘 ∈ 𝐴 ( - 1 · 𝐵 ) ) |
| 6 | 1 2 | fsumcl | ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 𝐵 ∈ ℂ ) |
| 7 | 6 | mulm1d | ⊢ ( 𝜑 → ( - 1 · Σ 𝑘 ∈ 𝐴 𝐵 ) = - Σ 𝑘 ∈ 𝐴 𝐵 ) |
| 8 | 2 | mulm1d | ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → ( - 1 · 𝐵 ) = - 𝐵 ) |
| 9 | 8 | sumeq2dv | ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 ( - 1 · 𝐵 ) = Σ 𝑘 ∈ 𝐴 - 𝐵 ) |
| 10 | 5 7 9 | 3eqtr3rd | ⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐴 - 𝐵 = - Σ 𝑘 ∈ 𝐴 𝐵 ) |