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	$$⊢\left(\phi \wedge j\in A\wedge k\in B\right)\to C=D$$
Assertion	2sumeq2dv	$$⊢\phi \to \sum _{j\in A}\sum _{k\in B}C=\sum _{j\in A}\sum _{k\in B}D$$

Proof

Step	Hyp	Ref	Expression
1		2sumeq2dv.1	$$⊢\left(\phi \wedge j\in A\wedge k\in B\right)\to C=D$$
2	1	3expa	$$⊢\left(\left(\phi \wedge j\in A\right)\wedge k\in B\right)\to C=D$$
3	2	sumeq2dv	$$⊢\left(\phi \wedge j\in A\right)\to \sum _{k\in B}C=\sum _{k\in B}D$$
4	3	sumeq2dv	$$⊢\phi \to \sum _{j\in A}\sum _{k\in B}C=\sum _{j\in A}\sum _{k\in B}D$$