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Metamath Proof Explorer

Theorem sumeq1d

Description: Equality deduction for sum. (Contributed by NM, 1-Nov-2005)

Ref Expression
Hypothesis sumeq1d.1 
|- ( ph -> A = B )
Assertion sumeq1d 
|- ( ph -> sum_ k e. A C = sum_ k e. B C )

Proof

Step Hyp Ref Expression
1 sumeq1d.1 
 |-  ( ph -> A = B )
2 sumeq1 
 |-  ( A = B -> sum_ k e. A C = sum_ k e. B C )
3 1 2 syl 
 |-  ( ph -> sum_ k e. A C = sum_ k e. B C )