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Proof step 11 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⌝ → lim(nσf) to n + 2 ⊑ nσf(n0), ∀ n. mon<(nσf(n)), ∀ n. nσf(n) ⊑ lim(nσf) ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 12:
⌜n⌝ + ⌜1⌝ < # lim(nσf), # lim(nσf) < ⌜n⌝ + ⌜2⌝ , mon(nσf), lim(nσf)[n + 1] ≤ lim(nσf)[n], ¬ # lim(nσf) < ⌜n⌝ + ⌜2⌝ → lim(nσf) to n + 2 ⊑ nσf(n0), ∀ n. nσf(n) ⊑ lim(nσf), ∀ n. mon<(nσf(n)) ⊦
Proof step 14:
⌜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. nσf(n) ⊑ lim(nσf), ∀ n. mon<(nσf(n)) ⊦ # lim(nσf) < ⌜n⌝ + ⌜2⌝

Rule argument

fma: # lim(nσf) < ⌜n⌝ + ⌜2⌝

Additional information

Used formulas: # lim(nσf) < ⌜n⌝ + ⌜2⌝