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Proof step 28 from proof rem-preserves-monls (previous step)

⌜n⌝ < # rem(⌜nl0⌝ , np), ⌜n + 1⌝ < # rem(⌜nl0⌝ , np), mon<(⌜nl0⌝ + ⌜m '⌝ ), rem(⌜nl0⌝ , np)[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)) ⊦ np(# nl0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
⌜n⌝ < # rem(⌜nl0⌝ , np), ⌜n⌝ + ⌜1⌝ < # rem(⌜nl0⌝ , np), mon<(⌜nl0⌝ + ⌜m '⌝ ), rem(⌜nl0⌝ , np)[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)) ⊦ np(# nl0)

Rule argument

Additional information

Used simplifier rules: 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)