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Proof step 3 from proof range-dups (previous step)

0 ≤ j, j < n→i(# ints), ∀ 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 = [], dups(ints)

User applied rule apply induction

And obtained the following new goals

Proof step 4:
0 ≤ j, j < n→i(# ints), ∀ 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) ⊦ i→n(j - 1) < i→n(j), ints = [], dups(ints)
Proof step 6:
0 ≤ j, j < n→i(# ints), ∀ 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) ⊦ (∀ i. i ∈ ints → 0 ≤ i ∧ i < j - 1) ∧ 0 ≤ j - 1 ∧ j - 1 < n→i(# ints), ints = [], dups(ints)

Rule argument

subst: [j0, ints] → [j - 1, ints]

Additional information

Used terms: j - 1, ints