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Proof step 5 from proof monls-from (previous step)

mon<(⌜nnf⌝ ) ⊦ ∀ n0. ⌜n0⌝ + ⌜1⌝ < # ⌜nnf from n⌝ → ⌜nnf from n⌝ [n0] < ⌜nnf from n⌝ [n0 + 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
mon<(⌜nnf⌝ ), nnf((n0 + 1) + n) ≤ nnf(n0 + n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ((nnf from n))(m) = nnf(m + n); (s)% ⊦ ⌜nnf⌝ [n] = nnf(n); (s)%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% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; (s)% ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)% ⊦ # ⌜nnf⌝ = ∞; (s)