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Proof step 24 from proof mininv-inverts-inner-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
⌜n⌝ + ⌜1⌝ = # nσ, nσ[n] < nσ[(# nσ).n - 1], nσ[n] < nσ[min(nσ, nσ[n] + 1)], mon<(nσ), min(nσ, nσ[n] + 1) ≤ n, ∀ m. ¬ n < m ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); (s)% ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s)%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% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ n∞ < n∞0 → (n∞ + ⌜1⌝ < n∞0 ↔ ¬(n∞ + ⌜1⌝ = n∞0 ∧ n∞ ≠ ∞)); (ss)