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Proof step 2 from proof sum-swap (previous step)

⊦ Σ(n, λ k0. Σ(0, λ k1. nnf(k0, k1))) = Σ(0, λ k1. Σ(n, λ k0. nnf(k0, k1)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ Σ(n, λ k. 0) = 0

Rule argument

Additional information

Used simplifier rules: sum-base: ⊦ Σ(0, nf) = 0;