Metamath Proof Explorer


Theorem sumeq1d

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

Ref Expression
Hypothesis sumeq1d.1 φ A = B
Assertion sumeq1d φ k A C = k B C

Proof

Step Hyp Ref Expression
1 sumeq1d.1 φ A = B
2 sumeq1 A = B k A C = k B C
3 1 2 syl φ k A C = k B C