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Proof step 21 from proof mininv-inverts-inner-succ (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: inst: name: monls-ls seq: ⊦ mon<(nσ) ∧ m < n ∧ ⌜n⌝ < # nσ → nσ[m] < nσ[n] subst: [nσ, m, n] → [nσ, min(nσ, nσ[n] + 1), n] flags: current, []

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%ler: ⊦ m ≤ m; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: nσ, min(nσ, nσ[n] + 1), n