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Proof step 15 from proof range-dups (previous step)

i + 1 < n→i(# ints), i ∈ ints, 0 ≤ i, i < n→i(# 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) ⊦ ints = [], # ints = 1, ~1 = i, dups(ints), (i0 ∈ ints -1l i → 0 ≤ i0 ∧ i0 < i) ∧ i < n→i(# ints - 1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
# ints = m + 1 ∧ # ints ≠ 0, i + 1 < n→i(# ints), i ∈ ints, 0 ≤ i, i < n→i(# 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) ⊦ ints = [], # ints = 1, ~1 = i, dups(ints), (i0 ∈ ints -1l i → 0 ≤ i0 ∧ i0 < i) ∧ i < n→i(m)

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# ints, m] flags:

Additional information

Used simplifier rules: l: ⊦ # ints = 0 ↔ ints = []; from specification list-data[intlista]%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # ints, m