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

∀ n0. ¬ m0 < n0 → nP0(n + n0) = nP1(n + n0), ∀ nP0, n, nP1. (∀ n0. n0 < m0 → nP0(n + n0) = nP1(n + n0)) → nP0 ⊗_(n, m0) = nP1 ⊗_(n, m0) ⊦ nP0(n) * nP0 ⊗_(n + 1, m0) = nP1 ⊗_(n, m0 + 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ nP ⊗_(n, m + 1) = nP(n) * nP ⊗_(n + 1, m)
Proof step 7:
∀ nP0, n, nP1. (∀ n0. n0 < m0 → nP0(n + n0) = nP1(n + n0)) → nP0 ⊗_(n, m0) = nP1 ⊗_(n, m0), ∀ n0. ¬ m0 < n0 → nP0(n + n0) = nP1(n + n0) ⊦ nP0(n) * nP0 ⊗_(n + 1, m0) = nP1(n) * nP1 ⊗_(n + 1, m0)

Rule argument

spec: <toplevel> name: Sep-rec seq: ⊦ nP ⊗_(n, m + 1) = nP(n) * nP ⊗_(n + 1, m) subst: [m, n, nP] → [m0, n, nP1] flags: <all paths>

Additional information

Used terms: m0, n, nP1