Metamath Proof Explorer


Theorem fsum2mul

Description: Separate the nested sum of the product C ( j ) x. D ( k ) . (Contributed by NM, 13-Nov-2005) (Revised by Mario Carneiro, 24-Apr-2014)

Ref Expression
Hypotheses fsum2mul.1 ( 𝜑𝐴 ∈ Fin )
fsum2mul.2 ( 𝜑𝐵 ∈ Fin )
fsum2mul.3 ( ( 𝜑𝑗𝐴 ) → 𝐶 ∈ ℂ )
fsum2mul.4 ( ( 𝜑𝑘𝐵 ) → 𝐷 ∈ ℂ )
Assertion fsum2mul ( 𝜑 → Σ 𝑗𝐴 Σ 𝑘𝐵 ( 𝐶 · 𝐷 ) = ( Σ 𝑗𝐴 𝐶 · Σ 𝑘𝐵 𝐷 ) )

Proof

Step Hyp Ref Expression
1 fsum2mul.1 ( 𝜑𝐴 ∈ Fin )
2 fsum2mul.2 ( 𝜑𝐵 ∈ Fin )
3 fsum2mul.3 ( ( 𝜑𝑗𝐴 ) → 𝐶 ∈ ℂ )
4 fsum2mul.4 ( ( 𝜑𝑘𝐵 ) → 𝐷 ∈ ℂ )
5 2 4 fsumcl ( 𝜑 → Σ 𝑘𝐵 𝐷 ∈ ℂ )
6 1 5 3 fsummulc1 ( 𝜑 → ( Σ 𝑗𝐴 𝐶 · Σ 𝑘𝐵 𝐷 ) = Σ 𝑗𝐴 ( 𝐶 · Σ 𝑘𝐵 𝐷 ) )
7 2 adantr ( ( 𝜑𝑗𝐴 ) → 𝐵 ∈ Fin )
8 4 adantlr ( ( ( 𝜑𝑗𝐴 ) ∧ 𝑘𝐵 ) → 𝐷 ∈ ℂ )
9 7 3 8 fsummulc2 ( ( 𝜑𝑗𝐴 ) → ( 𝐶 · Σ 𝑘𝐵 𝐷 ) = Σ 𝑘𝐵 ( 𝐶 · 𝐷 ) )
10 9 sumeq2dv ( 𝜑 → Σ 𝑗𝐴 ( 𝐶 · Σ 𝑘𝐵 𝐷 ) = Σ 𝑗𝐴 Σ 𝑘𝐵 ( 𝐶 · 𝐷 ) )
11 6 10 eqtr2d ( 𝜑 → Σ 𝑗𝐴 Σ 𝑘𝐵 ( 𝐶 · 𝐷 ) = ( Σ 𝑗𝐴 𝐶 · Σ 𝑘𝐵 𝐷 ) )