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Proof step 3 from proof stutf-slowincrease (previous step)

∀ nσ, m. ¬(nσ[m] + n0 < nσ[m + n0] ∧ ⌜m⌝ + ⌜n0⌝ < # nσ ∧ stutf(nσ)), nσ[m] + n0 +1 < nσ[m + n0 +1], ⌜m⌝ + ⌜n0 +1⌝ < # nσ, stutf(nσ) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⌜m⌝ + ⌜n0⌝ + ⌜1⌝ < # nσ, nσ[m] + n0 + 1 < nσ[m + n0 + 1], stutf(nσ), ∀ nσ0, m0. ¬ nσ0[m0] + n0 < nσ0[m0 + n0] ∨ ¬ ⌜m0⌝ + ⌜n0⌝ < # nσ0 ∨ ¬ stutf(nσ0) ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%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%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2