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Proof step 12 from proof range-dups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
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)

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%transitivity: ⊦ i < i0 ∧ i0 < i1 → i < i1; (forward) from specification oelem[intlista]%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%les: ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs