Metamath Proof Explorer


Theorem sumeq12rdv

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

Ref Expression
Hypotheses sumeq12rdv.1 φ A = B
sumeq12rdv.2 φ k B C = D
Assertion sumeq12rdv φ k A C = k B D

Proof

Step Hyp Ref Expression
1 sumeq12rdv.1 φ A = B
2 sumeq12rdv.2 φ k B C = D
3 1 sumeq1d φ k A C = k B C
4 2 sumeq2dv φ k B C = k B D
5 3 4 eqtrd φ k A C = k B D