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Proof step 7 from proof range-dups (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 8:
j = 0, j ≤ n→i(# ints), j < n→i(# ints), 0 ≤ 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 = [], dups(ints), 0 < j ∧ (i ∈ ints → 0 ≤ i ∧ i + 1 < j)
Proof step 10:
j ≤ n→i(# ints), j < n→i(# ints), 0 ≤ 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) ⊦ j = 0, ints = [], dups(ints), 0 < j ∧ (i ∈ ints → 0 ≤ i ∧ i + 1 < j)

Rule argument

fma: j = 0

Additional information

Used formulas: j = 0