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Proof step 4 from proof monls-append1 (previous step)

mon<(⌜nl⌝ + ⌜nl0⌝ ) ⊦ ∀ n. ⌜n⌝ + ⌜1⌝ < # ⌜nl⌝ → ⌜nl⌝ [n] < ⌜nl⌝ [n + 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
nl[n + 1] ≤ nl[n], 1 + n < # nl, mon<(⌜nl⌝ + ⌜nl0⌝ ) ⊦ nl = []

Rule argument

Additional information

Used simplifier rules: l: ⊦ # nl = 0 ↔ nl = []; from specification list-data[natstream]%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%size-fin: ⊦ # ⌜nl⌝ = ⌜# nl⌝ ; from specification stream[natstream]