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

∀ n. (λ n. emp)⊗_(n, m0) = emp ⊦ (λ n. emp)⊗_(n, m0 + 1) = emp

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:
∀ n. (λ n. emp)⊗_(n, m0) = emp ⊦ emp * (λ n. emp)⊗_(n + 1, m0) = emp

Rule argument

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

Additional information

Used terms: m0, n, λ n. emp