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Proof step 1 from proof sum-append (conclusion)

⊦ Σ(nats + nats0) = Σ nats + Σ nats0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ nats0. Σ(nats1 + nats0) = Σ nats1 + Σ nats0 ⊦ Σ((m + nats1) + nats0) = Σ(m + nats1) + Σ nats0

Rule argument

term: nats

Additional information

Used simplifier rules: sum-empty: ⊦ Σ [] = 0;%natlist freely generated by [], +;;% ⊦ [] + nats = nats;% ⊦ (nats + nats0) + nats1 = nats + nats0 + nats1;% ⊦ 0 + m = m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;