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

∀ nf0, nf1. Σ(n0, λ k. nf0(k) + nf1(k)) = Σ(n0, nf0) + Σ(n0, nf1) ⊦ Σ(n0 +1, λ k. nf0(k) + nf1(k)) = Σ(n0 +1, nf0) + Σ(n0 +1, nf1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ nf0, nf1. Σ(n0, λ k. nf0(k) + nf1(k)) = Σ(n0, nf0) + Σ(n0, nf1) ⊦ Σ(n0, λ k. nf0(k) + nf1(k)) = Σ(n0, nf0) + Σ(n0, nf1)

Rule argument

Additional information

Used simplifier rules: sum-rec: ⊦ Σ(n + 1, nf) = nf(n) + Σ(n, nf);%add-c: ⊦ i + j = j + i; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%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