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Proof step 5 from proof Sep-split (previous step)

∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦ nP ⊗_(n, n0 + m +1) = nP ⊗_(n, m +1) * nP ⊗_(n + m +1, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦ nP ⊗_(n, n0 + m + 1) = nP ⊗_(n, m + 1) * nP ⊗_(n + m + 1, n0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)