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Proof step 13 from proof monls-mon-implies-monls (previous step)

⌜n⌝ + ⌜1⌝ < # lim(nσf), # lim(nσf) < ⌜n⌝ + ⌜2⌝ , mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], ∀ n. nσf(n) ⊑ lim(nσf), ∀ n. mon<(nσf(n)) ⊦

User applied rule structural induction

And closed the goal!

Rule argument

term: # lim(nσf)

Additional information

Used simplifier rules: natinf freely generated by ∞, ⌜ ⌝;;% ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; (s)%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf% ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)% ⊦ ¬ ∞ < n∞; (s)