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Proof step 24 from proof rem-preserves-monls (previous step)

⌜n⌝ < # rem(⌜nl0⌝ , np) → (∃ m0. m0 < # nl0 ∧ nl0[m0] = rem(⌜nl0⌝ , np)[n] ∧ ¬ np(m0)), ⌜n⌝ + ⌜1⌝ = # rem(⌜nl0⌝ , np), mon<(⌜nl0⌝ + ⌜m '⌝ ), m ≤ rem(⌜nl0⌝ , np)[n], ∀ n0. n0 < n → rem(⌜nl0⌝ , np)[n0] < rem(⌜nl0⌝ , np)[n0 + 1], ∀ nl1, np. # nl1 < #(nl0 + m ') → mon<(⌜nl1⌝ ) → mon<(rem(⌜nl1⌝ , np)) ⊦ np(# nl0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
nl0[m0] = rem(⌜nl0⌝ , np)[n], ⌜n⌝ + ⌜1⌝ = # rem(⌜nl0⌝ , np), m0 < # nl0, m ≤ nl0[m0], mon<(⌜nl0⌝ + ⌜m '⌝ ), ∀ n0. n0 < n → rem(⌜nl0⌝ , np)[n0] < rem(⌜nl0⌝ , np)[n0 + 1], ∀ nl1, np. # nl1 < #(nl0 + m ') → mon<(⌜nl1⌝ ) → mon<(rem(⌜nl1⌝ , np)) ⊦ nl0 = [], np(m0), np(# nl0)

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%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%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