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	|- ( ( ph /\ j e. A /\ k e. B ) -> C = D )
Assertion	2sumeq2dv	|- ( ph -> sum_ j e. A sum_ k e. B C = sum_ j e. A sum_ k e. B D )

Proof

Step	Hyp	Ref	Expression
1		2sumeq2dv.1	|- ( ( ph /\ j e. A /\ k e. B ) -> C = D )
2	1	3expa	|- ( ( ( ph /\ j e. A ) /\ k e. B ) -> C = D )
3	2	sumeq2dv	|- ( ( ph /\ j e. A ) -> sum_ k e. B C = sum_ k e. B D )
4	3	sumeq2dv	|- ( ph -> sum_ j e. A sum_ k e. B C = sum_ j e. A sum_ k e. B D )