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Proof step 51 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, stutf(nσ), ⌜nσ[n1]⌝ < leninv(nσ), ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + 1] = nσ[n1] + 1, nσ[n1 + 1] = nσ[n1]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 52:
⊦ stutf(nσ) ↔ nσ ≠ [] ∧ nσ[0] = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1)
Proof step 53:
# nσ = ∞, nσ ≠ [] ∧ nσ[0] = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1), ⌜nσ[n1]⌝ < leninv(nσ), ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + 1] = nσ[n1] + 1, nσ[n1 + 1] = nσ[n1]

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