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Proof step 10 from proof monls-mon-implies-monls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
⌜n⌝ + ⌜1⌝ < # lim(nσf), mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], ¬ # lim(nσf) < ⌜n⌝ + ⌜2⌝ → lim(nσf) to n + 2 ⊑ nσf(n0), ∀ n. mon<(nσf(n)), ∀ n. nσf(n) ⊑ lim(nσf) ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)