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Proof step 22 from proof rem-preserves-monls (previous step)

⌜n⌝ < # rem(⌜nl0⌝ , np), mon<(⌜nl0⌝ + ⌜m '⌝ ), (rem(⌜nl0⌝ , np) + ⌜m '⌝ )[n + 1] ≤ rem(⌜nl0⌝ , np)[n], ∀ n. ⌜n⌝ + ⌜1⌝ < # rem(⌜nl0⌝ , np) → rem(⌜nl0⌝ , np)[n] < rem(⌜nl0⌝ , np)[n + 1], ∀ nl1, np. # nl1 < #(nl0 + m ') → mon<(⌜nl1⌝ ) → mon<(rem(⌜nl1⌝ , np)) ⊦ ⌜n + 1⌝ < # rem(⌜nl0⌝ , np), np(# nl0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m '[0] = m; (s)% ⊦ m - m = 0; (s)% ⊦ m + n - n + n1 = m - n1; (s)% ⊦ ⌜m⌝ .n = m; (s)%n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); (s)% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (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∞0 → (n∞ + ⌜1⌝ < n∞0 ↔ ¬(n∞ + ⌜1⌝ = n∞0 ∧ n∞ ≠ ∞)); (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%append-assoc: ⊦ (nσ + nσ0) + nσ1 = nσ + nσ0 + nσ1; from specification stream[natstream]% ⊦ ⌜nl⌝ [n] = nl[n]; (s)% ⊦ ¬ ⌜m⌝ < # nσ → (nσ + nσ0)[m] = nσ0[m - (# nσ).n]; (s)