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Proof step 21 from proof nodups-interval-size (previous step)

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))

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: e: ⊦ i ' ≠ []; from specification list[intlista]%len-02: ⊦ # i ' = 0 +1; from specification list[intlista]