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Proof step 15 from proof nodups-interval-size (previous step)

j1 ∈ ints -1l max(ints), max(ints) < n→i(# ints) + j0, j0 < j1, max(ints) ≤ j1, j0 < max(ints), j1 < n→i(# ints) + j0, j1 ∈ ints, ∀ i. i ∈ ints → j0 < i ∧ i ≤ j1, ∀ ints0, j0, j1. # ints0 < # ints → ¬ dups(ints0) ∧ (∀ i. i ∈ ints0 → j0 < i ∧ i ≤ j1) ∧ ints0 ≠ [] → n→i(# ints0) ≤ j1 - j0 ⊦ ints = max(ints) ', ints = [], dups(ints -1l max(ints)), dups(ints)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: list-del inst: intlista name: del1-in seq: ⊦ i ≠ i0 → (i ∈ ints -1l i0 ↔ i ∈ ints) subst: [i0, ints, i] → [max(ints), ints, j1] flags: rewrite

Additional information

Used simplifier rules: max-in-not-ls: ⊦ i ∈ ints → ¬ max(ints) < i;%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%dups-01: ⊦ ¬ dups(ints) → ¬ i ∈ ints -1l i; from specification list-del[intlista] Used terms: max(ints), ints, j1