Metamath Proof Explorer


Theorem sumeq1d

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

Ref Expression
Hypothesis sumeq1d.1 φA=B
Assertion sumeq1d φkAC=kBC

Proof

Step Hyp Ref Expression
1 sumeq1d.1 φA=B
2 sumeq1 A=BkAC=kBC
3 1 2 syl φkAC=kBC