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

⌜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)

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 23:
⊦ ⌜m⌝ < # rem(⌜nl⌝ , np) → (∃ m0. m0 < # nl ∧ nl[m0] = rem(⌜nl⌝ , np)[m] ∧ ¬ np(m0))
Proof step 24:
⌜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)

Rule argument

spec: mon-lim-filter-count inst: natstream name: rem-member1-fin seq: ⊦ ⌜m⌝ < # rem(⌜nl⌝ , np) → (∃ m0. m0 < # nl ∧ nl[m0] = rem(⌜nl⌝ , np)[m] ∧ ¬ np(m0)) subst: [m, nl, np] → [n, nl0, np] flags: []

Additional information

Used terms: n, nl0, np