Metamath Proof Explorer

Theorem sumeq2dv

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

Ref Expression
Hypothesis sumeq2dv.1 
|- ( ( ph /\ k e. A ) -> B = C )
Assertion sumeq2dv 
|- ( ph -> sum_ k e. A B = sum_ k e. A C )

Proof

Step Hyp Ref Expression
1 sumeq2dv.1 
 |-  ( ( ph /\ k e. A ) -> B = C )
2 1 ralrimiva 
 |-  ( ph -> A. k e. A B = C )
3 2 sumeq2d 
 |-  ( ph -> sum_ k e. A B = sum_ k e. A C )