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Proof step 10 from proof stutf-alltaken (previous step)

nσ[m0] < nσ[m0 + 1], nσ[m0] < n, ⌜m0⌝ < # nσ, stutf(nσ), ⌜m0⌝ + ⌜1⌝ < # nσ, n ≤ nσ[m0 + 1], m0 ≤ nσ[m0] → m ≤ nσ[m] ∧ nσ[m] = m0 ⊦ n = 0, nσ[m0 + 1] = 0, ∃ m. m ≤ m0 + 1 ∧ nσ[m] = n

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ stutf(nσ) ↔ nσ ≠ [] ∧ nσ[0] = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1)
Proof step 12:
nσ[m0] < nσ[m0 + 1], nσ[m0] < n, ⌜m0⌝ < # nσ, nσ ≠ [] ∧ nσ[0] = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1), ⌜m0⌝ + ⌜1⌝ < # nσ, n ≤ nσ[m0 + 1], m0 ≤ nσ[m0] → m ≤ nσ[m] ∧ nσ[m] = m0 ⊦ n = 0, nσ[m0 + 1] = 0, ∃ m. m ≤ m0 + 1 ∧ nσ[m] = n

Rule argument

spec: inst: 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: rewrite, current

Additional information

Used terms: nσ