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Proof step 8 from proof monle-leq (previous step)

⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, nσ[(m1 + 1) + m] < nσ[m], mon≤(nσ), m1 ≠ m1 + m - m ∨ ¬ nσ[m1 + m] < nσ[m] ∨ ¬ ⌜m1 + m⌝ < # nσ ∨ ¬ mon≤(nσ) ∨ ¬ m ≤ m1 + m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, nσ[(m1 + 1) + m] < nσ[m], ⌜m1⌝ + ⌜m⌝ < # nσ, nσ[(m1 + 1) + m] < nσ[m1 + m], mon≤(nσ), nσ[m] ≤ nσ[m1 + m] ⊦

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%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 + n < m; (s)% ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s)% ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (s)%lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n - m = n; (s)