Metamath Proof Explorer


Theorem sumeq12dv

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

Ref Expression
Hypotheses sumeq12dv.1 φ A = B
sumeq12dv.2 φ k A C = D
Assertion sumeq12dv φ k A C = k B D

Proof

Step Hyp Ref Expression
1 sumeq12dv.1 φ A = B
2 sumeq12dv.2 φ k A C = D
3 2 sumeq2dv φ k A C = k A D
4 1 sumeq1d φ k A D = k B D
5 3 4 eqtrd φ k A C = k B D