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 )