Metamath Proof Explorer

Theorem sumeq2sdv

Description: Equality deduction for sum. (Contributed by NM, 3-Jan-2006) (Proof shortened by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypothesis sumeq2sdv.1 
|- ( ph -> B = C )
Assertion sumeq2sdv 
|- ( ph -> sum_ k e. A B = sum_ k e. A C )

Proof

Step Hyp Ref Expression
1 sumeq2sdv.1 
 |-  ( ph -> B = C )
2 1 ralrimivw 
 |-  ( ph -> A. k e. A B = C )
3 2 sumeq2d 
 |-  ( ph -> sum_ k e. A B = sum_ k e. A C )