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Proof step 7 from proof monle-leq (previous step)

nσ[(m1 + 1) + m] < nσ[m], mon≤(nσ), ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, m1 = m1 + m - m ∧ mon≤(nσ) ∧ m ≤ m1 + m ∧ ⌜m1 + m⌝ < # nσ ∧ nσ[m1 + m] < nσ[m] → nσ[m] = 0 ⊦ nσ[m] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%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%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%add-ls-cancel-01: ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; from specification nat-inf%lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%lep: ⊦ m ≤ m + n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n - m = n; from specification nat