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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	⊢ ( 𝜑 → 𝐴 ∈ Fin )
		fsum2mul.2	⊢ ( 𝜑 → 𝐵 ∈ Fin )
		fsum2mul.3	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) → 𝐶 ∈ ℂ )
		fsum2mul.4	⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐵 ) → 𝐷 ∈ ℂ )
	Assertion	fsum2mul	⊢ ( 𝜑 → Σ 𝑗 ∈ 𝐴 Σ 𝑘 ∈ 𝐵 ( 𝐶 · 𝐷 ) = ( Σ 𝑗 ∈ 𝐴 𝐶 · Σ 𝑘 ∈ 𝐵 𝐷 ) )

Proof

Step	Hyp	Ref	Expression
1		fsum2mul.1	⊢ ( 𝜑 → 𝐴 ∈ Fin )
2		fsum2mul.2	⊢ ( 𝜑 → 𝐵 ∈ Fin )
3		fsum2mul.3	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) → 𝐶 ∈ ℂ )
4		fsum2mul.4	⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐵 ) → 𝐷 ∈ ℂ )
5	2 4	fsumcl	⊢ ( 𝜑 → Σ 𝑘 ∈ 𝐵 𝐷 ∈ ℂ )
6	1 5 3	fsummulc1	⊢ ( 𝜑 → ( Σ 𝑗 ∈ 𝐴 𝐶 · Σ 𝑘 ∈ 𝐵 𝐷 ) = Σ 𝑗 ∈ 𝐴 ( 𝐶 · Σ 𝑘 ∈ 𝐵 𝐷 ) )
7	2	adantr	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) → 𝐵 ∈ Fin )
8	4	adantlr	⊢ ( ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) ∧ 𝑘 ∈ 𝐵 ) → 𝐷 ∈ ℂ )
9	7 3 8	fsummulc2	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) → ( 𝐶 · Σ 𝑘 ∈ 𝐵 𝐷 ) = Σ 𝑘 ∈ 𝐵 ( 𝐶 · 𝐷 ) )
10	9	sumeq2dv	⊢ ( 𝜑 → Σ 𝑗 ∈ 𝐴 ( 𝐶 · Σ 𝑘 ∈ 𝐵 𝐷 ) = Σ 𝑗 ∈ 𝐴 Σ 𝑘 ∈ 𝐵 ( 𝐶 · 𝐷 ) )
11	6 10	eqtr2d	⊢ ( 𝜑 → Σ 𝑗 ∈ 𝐴 Σ 𝑘 ∈ 𝐵 ( 𝐶 · 𝐷 ) = ( Σ 𝑗 ∈ 𝐴 𝐶 · Σ 𝑘 ∈ 𝐵 𝐷 ) )