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Proof step 18 from proof range-dups (previous step)

# ints = m + 1, i0 ∈ ints, i0 ∈ ints -1l i, i < n→i(m), i ∈ ints, 0 ≤ i, i ≤ n→i(m), ∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# 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

Rule argument

Additional information

Used simplifier rules: del1-in: ⊦ i ≠ i0 → (i ∈ ints -1l i0 ↔ i ∈ ints); from specification list-del[intlista]