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Proof step 6 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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ nP ⊗_(n, m + 1) = nP(n) * nP ⊗_(n + 1, m)
Proof step 8:
∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦ nP(n) * nP ⊗_(n + 1, m + n0) = nP ⊗_(n, m + 1) * nP ⊗_(n + m + 1, n0)

Rule argument

spec: <toplevel> name: Sep-rec seq: ⊦ nP ⊗_(n, m + 1) = nP(n) * nP ⊗_(n + 1, m) subst: [m, n, nP] → [m + n0, n, nP] flags: <all paths>

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m + n0, n, nP