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Proof step 13 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)

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [j0, ints] → [i, ints -1l i]

Additional information

Used simplifier rules: dups: ⊦ dups(ints -1l i) → dups(ints); (forward) from specification list-del[intlista]%rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1; from specification intnat%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%irreflexivity: ⊦ ¬ i < i; from specification oelem[intlista]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: i, ints -1l i