Metamath Proof Explorer


Theorem isummulc1

Description: An infinite sum multiplied by a constant. (Contributed by NM, 13-Nov-2005) (Revised by Mario Carneiro, 23-Apr-2014)

Ref Expression
Hypotheses isumcl.1 𝑍 = ( ℤ𝑀 )
isumcl.2 ( 𝜑𝑀 ∈ ℤ )
isumcl.3 ( ( 𝜑𝑘𝑍 ) → ( 𝐹𝑘 ) = 𝐴 )
isumcl.4 ( ( 𝜑𝑘𝑍 ) → 𝐴 ∈ ℂ )
isumcl.5 ( 𝜑 → seq 𝑀 ( + , 𝐹 ) ∈ dom ⇝ )
summulc.6 ( 𝜑𝐵 ∈ ℂ )
Assertion isummulc1 ( 𝜑 → ( Σ 𝑘𝑍 𝐴 · 𝐵 ) = Σ 𝑘𝑍 ( 𝐴 · 𝐵 ) )

Proof

Step Hyp Ref Expression
1 isumcl.1 𝑍 = ( ℤ𝑀 )
2 isumcl.2 ( 𝜑𝑀 ∈ ℤ )
3 isumcl.3 ( ( 𝜑𝑘𝑍 ) → ( 𝐹𝑘 ) = 𝐴 )
4 isumcl.4 ( ( 𝜑𝑘𝑍 ) → 𝐴 ∈ ℂ )
5 isumcl.5 ( 𝜑 → seq 𝑀 ( + , 𝐹 ) ∈ dom ⇝ )
6 summulc.6 ( 𝜑𝐵 ∈ ℂ )
7 1 2 3 4 5 6 isummulc2 ( 𝜑 → ( 𝐵 · Σ 𝑘𝑍 𝐴 ) = Σ 𝑘𝑍 ( 𝐵 · 𝐴 ) )
8 1 2 3 4 5 isumcl ( 𝜑 → Σ 𝑘𝑍 𝐴 ∈ ℂ )
9 8 6 mulcomd ( 𝜑 → ( Σ 𝑘𝑍 𝐴 · 𝐵 ) = ( 𝐵 · Σ 𝑘𝑍 𝐴 ) )
10 6 adantr ( ( 𝜑𝑘𝑍 ) → 𝐵 ∈ ℂ )
11 4 10 mulcomd ( ( 𝜑𝑘𝑍 ) → ( 𝐴 · 𝐵 ) = ( 𝐵 · 𝐴 ) )
12 11 sumeq2dv ( 𝜑 → Σ 𝑘𝑍 ( 𝐴 · 𝐵 ) = Σ 𝑘𝑍 ( 𝐵 · 𝐴 ) )
13 7 9 12 3eqtr4d ( 𝜑 → ( Σ 𝑘𝑍 𝐴 · 𝐵 ) = Σ 𝑘𝑍 ( 𝐴 · 𝐵 ) )