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Proof step 21 from proof rem-preserves-monls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: one-03: ⊦ m '[0] = m; from specification list-del[natstream]%zero: ⊦ m - m = 0; from specification nat%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat%sel: ⊦ ⌜m⌝ .n = m; from specification nat-inf-basic%prime-add-fin: n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%eqls-succ: ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); from specification nat-inf%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%inf-inseq-reduce: ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); from specification nat-inf%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-notlssucc: ⊦ ⌜m⌝ < n∞ → (¬ ⌜m⌝ + ⌜1⌝ < n∞ ↔ ⌜m⌝ + ⌜1⌝ = n∞); from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%append-assoc: ⊦ (nσ + nσ0) + nσ1 = nσ + nσ0 + nσ1; from specification stream[natstream]%sel-fin: ⊦ ⌜nl⌝ [n] = nl[n]; from specification stream[natstream]%get-append-notls: ⊦ ¬ ⌜m⌝ < # nσ → (nσ + nσ0)[m] = nσ0[m - (# nσ).n]; from specification stream[natstream]