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Proof step 19 from proof range-dups (previous step)

# ints = m + 1, i < n→i(m), i ∈ ints, 0 ≤ i, i ≤ n→i(m), i0 ∈ ints, ∀ i0. i0 ∈ ints → 0 ≤ i0 ∧ i0 ≤ i, ∀ j0, ints. i→n(j0) < i→n(i + 1) → (∀ i. i ∈ ints → 0 ≤ i ∧ i < j0) ∧ 0 ≤ j0 ∧ j0 < n→i(# ints) → dups(ints) ⊦ i0 = i, m = 0, ~1 = i, ints = [], dups(ints), 0 ≤ i0 ∧ i0 < i

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: i0 abort: F discard: F computed: T

Additional information

Used simplifier rules: lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: i0