Metamath Proof Explorer


Theorem sumeq12rdv

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

Ref Expression
Hypotheses sumeq12rdv.1 φA=B
sumeq12rdv.2 φkBC=D
Assertion sumeq12rdv φkAC=kBD

Proof

Step Hyp Ref Expression
1 sumeq12rdv.1 φA=B
2 sumeq12rdv.2 φkBC=D
3 1 sumeq1d φkAC=kBC
4 2 sumeq2dv φkBC=kBD
5 3 4 eqtrd φkAC=kBD