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Proof step 64 from proof inv-get-inverse-between (previous step)

mon≤(nσ) ∧ n1 + m + 1 ≤ m1 ∧ ⌜m1⌝ < # nσ → nσ[n1 + m + 1] ≤ nσ[m1], # nσ = ∞, nσ[m1 + 1] = nσ[n1] + 1, n1 + m < m1, nσ[m1] ≤ nσ[n1] + 1, ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), nσ[n1] ≤ nσ[n1 + m + 1], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. nσ[m] = nσ[n1] + 1 → ¬ m1 + 1 < m, ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ m1 = 0, nσ[n1 + m + 1] = nσ[n1] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%ler: ⊦ m ≤ m; from specification nat%lep: ⊦ m ≤ m + n; from specification nat%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat