Theorem 2sumeq2dv

Description: Equality deduction for double sum. (Contributed by NM, 3-Jan-2006) (Revised by Mario Carneiro, 31-Jan-2014)

	Ref	Expression
Hypothesis	2sumeq2dv.1	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵 ) → 𝐶 = 𝐷 )
Assertion	2sumeq2dv	⊢ ( 𝜑 → Σ 𝑗 ∈ 𝐴 Σ 𝑘 ∈ 𝐵 𝐶 = Σ 𝑗 ∈ 𝐴 Σ 𝑘 ∈ 𝐵 𝐷 )

Proof

Step	Hyp	Ref	Expression
1		2sumeq2dv.1	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ∧ 𝑘 ∈ 𝐵 ) → 𝐶 = 𝐷 )
2	1	3expa	⊢ ( ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) ∧ 𝑘 ∈ 𝐵 ) → 𝐶 = 𝐷 )
3	2	sumeq2dv	⊢ ( ( 𝜑 ∧ 𝑗 ∈ 𝐴 ) → Σ 𝑘 ∈ 𝐵 𝐶 = Σ 𝑘 ∈ 𝐵 𝐷 )
4	3	sumeq2dv	⊢ ( 𝜑 → Σ 𝑗 ∈ 𝐴 Σ 𝑘 ∈ 𝐵 𝐶 = Σ 𝑗 ∈ 𝐴 Σ 𝑘 ∈ 𝐵 𝐷 )