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Proof step 23 from proof mininv-inverts-inner-succ (previous step)

⌜min(nσ, nσ[n] + 1)⌝ < # nσ, nσ[n] < nσ[min(nσ, nσ[n] + 1)], ⌜n⌝ < # nσ, mon<(nσ), # nσ = ∞ ∨ nσ[n] < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m, ⌜n⌝ + ⌜1⌝ < # nσ ∧ nσ[n] < nσ[n + 1] → ¬ n < min(nσ, nσ[n] + 1) ⊦ ⌜n⌝ + ⌜1⌝ < # nσ, min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
⌜n⌝ + ⌜1⌝ = # nσ, nσ[n] < nσ[min(nσ, nσ[n] + 1)], mon<(nσ), min(nσ, nσ[n] + 1) ≤ n, nσ[n] < nσ[(# nσ).n - 1], ∀ m. ¬ n < m ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0, nσ[(# nσ).n - 1] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%eqls-succ: ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); from specification nat-inf%sum-notinf: ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); from specification nat-inf%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%inf-inseq-reduce: ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%ls-notlssucc: ⊦ ⌜m⌝ < n∞ → (¬ ⌜m⌝ + ⌜1⌝ < n∞ ↔ ⌜m⌝ + ⌜1⌝ = n∞); from specification nat-inf