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Proof step 8 from proof Sep-split (previous step)

∀ 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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ nP ⊗_(n, m + 1) = nP(n) * nP ⊗_(n + 1, m)
Proof step 10:
∀ nP, n, n0. nP ⊗_(n, n0 + m) = nP ⊗_(n, m) * nP ⊗_(n + m, n0) ⊦ nP(n) * nP ⊗_(n + 1, m + n0) = (nP(n) * nP ⊗_(n + 1, m)) * 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, n, nP] flags: <all paths>

Additional information

Used terms: m, n, nP