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Proof step 11 from proof monls-mon-implies-monls (previous step)

# lim(nσf) < ⌜n⌝ + ⌜2⌝ , ⌜n⌝ + ⌜1⌝ < # lim(nσf), mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], ∀ n. mon<(nσf(n)), ∀ n. nσf(n) ⊑ lim(nσf), ¬ # lim(nσf) < ⌜n⌝ + ⌜2⌝ → lim(nσf) to n + 2 ⊑ nσf(n0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: 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%ref-1: ⊦ ¬ n < n; from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%less-succ-leq: ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; from specification nat-inf%add-ls-cancel-num-04: ⊦ m ≤ n → (⌜(* m)⌝ + n∞2 < ⌜(* n)⌝ + n∞1 ↔ n∞2 < ⌜n - m⌝ + n∞1); from specification nat-inf%ls-trans: ⊦ n∞ < n∞1 ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf