This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

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	$⊢ (φ \land j \in A \land k \in B) \to C = D$
	Assertion	2sumeq2dv	$⊢ φ \to \sum_{j \in A} \sum_{k \in B} C = \sum_{j \in A} \sum_{k \in B} D$

Step	Hyp	Ref	Expression
1		2sumeq2dv.1	$⊢ (φ \land j \in A \land k \in B) \to C = D$
2	1	3expa	$⊢ ((φ \land j \in A) \land k \in B) \to C = D$
3	2	sumeq2dv	$⊢ (φ \land j \in A) \to \sum_{k \in B} C = \sum_{k \in B} D$
4	3	sumeq2dv	$⊢ φ \to \sum_{j \in A} \sum_{k \in B} C = \sum_{j \in A} \sum_{k \in B} D$