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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false); (s)%m < n ⊦ (nnf to n)[m] = nnf(m); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ⌜nl⌝ [n] = nl[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% ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s)% ⊦ # ⌜nnf to n⌝ = ⌜n⌝ ; (s)