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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