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Proof step 4 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 5:
mon<(⌜nnf⌝ ), 1 + n0 < n, nnf(n0 + 1) ≤ nnf(n0) ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%get-infto: m < n ⊦ (nnf to n)[m] = nnf(m); from specification stream[natstream]%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%sel-fin: ⊦ ⌜nl⌝ [n] = nl[n]; from specification stream[natstream]%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%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%inf-size-to: ⊦ # ⌜nnf to n⌝ = ⌜n⌝ ; from specification stream[natstream]