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Proof step 11 from proof Sep-split (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ m + n = n + m
Proof step 13:
nP ⊗_(n + 1, m + n0) = nP ⊗_(n + 1, m) * nP ⊗_((n + 1) + m, n0) ⊦ nP(n) * nP ⊗_(n + 1, m + n0) = (nP(n) * nP ⊗_(n + 1, m)) * nP ⊗_(n + m + 1, n0)

Rule argument

spec: nat-basic name: c seq: ⊦ m + n = n + m subst: [n, m] → [m, n0] flags: <all paths>

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic Used terms: m, n0