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Proof step 5 from proof sum-swap (previous step)

∀ n, nnf. Σ(n, λ k0. Σ(m0, λ k1. nnf(k0, k1))) = Σ(m0, λ k1. Σ(n, λ k0. nnf(k0, k1))) ⊦ Σ(n, λ k0. Σ(m0 +1, λ k1. nnf(k0, k1))) = Σ(m0 +1, λ k1. Σ(n, λ k0. nnf(k0, k1)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ n, nnf. Σ(n, λ k0. Σ(m0, λ k1. nnf(k0, k1))) = Σ(m0, λ k1. Σ(n, λ k0. nnf(k0, k1))) ⊦ Σ(n, λ k. nnf(k, m0) + Σ(m0, λ k1. nnf(k, k1))) = Σ(n, λ k0. nnf(k0, m0)) + Σ(m0, λ k1. Σ(n, λ k0. nnf(k0, k1)))

Rule argument

Additional information

Used simplifier rules: sum-rec: ⊦ Σ(n + 1, nf) = nf(n) + Σ(n, nf);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2