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Proof step 4 from proof stutf-notbig (previous step)

n0 + 1 < nσ[n0 + 1], stutf(nσ), ⌜n0⌝ + ⌜1⌝ < # nσ, stutf(nσ) ∧ ⌜n0⌝ < # nσ → ¬ n0 < nσ[n0] ⊦ nσ[n0 + 1] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⌜n0⌝ + ⌜1⌝ < # nσ, stutf(nσ), n0 + 1 < nσ[n0 + 1], ⌜n0⌝ < # nσ, nσ[n0] ≤ n0 ⊦ nσ[n0 + 1] = 0

Rule argument

Additional information

Used simplifier rules: 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%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%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%lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf