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

∀ nats0. Σ(nats1 + nats0) = Σ nats1 + Σ nats0 ⊦ Σ((m + nats1) + nats0) = Σ(m + nats1) + Σ nats0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ nats0. Σ(nats1 + nats0) = Σ nats1 + Σ nats0 ⊦ Σ(nats1 + nats0) = Σ nats0 + Σ nats1

Rule argument

Additional information

Used simplifier rules: sum-rec: ⊦ Σ(n ' + nats) = n + Σ nats;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (nats + nats0) + nats1 = nats + nats0 + nats1;% ⊦ m + nats = m ' + nats;