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Proof step 20 from proof nodups-interval-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
ints = max(ints) ', j1 < n→i(# ints) + j0, j0 < max(ints), max(ints) ≤ j1, j0 < j1, max(ints) < n→i(# ints) + j0, ∀ 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 = [], # ints = 1, dups(ints), dups(ints -1l max(ints))

Rule argument

Additional information

Used simplifier rules: l: ⊦ # ints = 0 ↔ ints = []; from specification list-data[intlista]%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat