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Proof step 6 from proof stutf-monle (previous step)

nσ[n + 1] < nσ[n], ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ stutf(nσ) ↔ nσ ≠ [] ∧ nσ[0] = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1)
Proof step 8:
nσ[0] = 0, ⌜n⌝ + ⌜1⌝ < # nσ, nσ[n + 1] < nσ[n], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1 ⊦ nσ = []

Rule argument

spec: <toplevel> name: stutf-def seq: ⊦ stutf(nσ) ↔ nσ ≠ [] ∧ nσ[0] = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1) subst: [nσ] → [nσ] flags: <all paths>

Additional information

Used terms: nσ