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Proof step 136 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 137:
# nσ = ∞, nσ[n] = nσ[n1] + 1, n1 < m1, n1 + m + 1 < m1, n ≤ m1, stutf(nσ), nσ[n1] < nσ[m1], n1 + m + 1 < n, nσ[n] ≤ nσ[m1], nσ[n1 + 1] ≤ nσ[n1 + m + 1], ∀ m. nσ[m] = nσ[n] → ¬ n < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + 1] = nσ[n1], nσ[n] = nσ[m1] + 1, nσ[m1] = 0, nσ[n1 + m + 1] = nσ[n], n = 0, m1 = 0

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%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%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%addeqinf-cancel: ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; from specification nat-inf%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; from specification nat-inf%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat