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Proof step 3 from proof Sep-range (previous step)

∀ nP, n. nP ⊗_(n, m0) = (λ n0. ⌜n0 < n + m0⌝ → nP(n0))⊗_(n, m0) ⊦ nP ⊗_(n, m0 + 1) = (λ n0. ⌜¬ n + m0 < n0⌝ → nP(n0))⊗_(n, m0 + 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ nP ⊗_(n, m + 1) = nP(n) * nP ⊗_(n + 1, m)
Proof step 5:
∀ nP, n. nP ⊗_(n, m0) = (λ n0. ⌜n0 < n + m0⌝ → nP(n0))⊗_(n, m0) ⊦ nP(n) * nP ⊗_(n + 1, m0) = (λ n0. ⌜¬ n + m0 < n0⌝ → nP(n0))⊗_(n, m0 + 1)

Rule argument

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

Additional information

Used terms: m0, n, nP