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Proof step 14 from proof count-i2j-no-dups (previous step)

i1 + i = i2 + 1, dups(count-i2j(i, i2) + (i2 + 1)'), 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ ¬ i ∈ ints → (dups(ints + i ') ↔ dups(ints))
Proof step 16:
¬ ¬ i2 + 1 ∈ count-i2j(i, i2), i1 + i = i2 + 1, dups(count-i2j(i, i2) + (i2 + 1)'), 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦
Proof step 17:
¬ i2 + 1 ∈ count-i2j(i, i2), i1 + i = i2 + 1, dups(count-i2j(i, i2)), 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦

Rule argument

spec: list-dup inst: intlista name: l-03 seq: ⊦ ¬ i ∈ ints → (dups(ints + i ') ↔ dups(ints)) subst: [i, ints] → [i2 + 1, count-i2j(i, i2)] flags: rewrite

Additional information

Used terms: i2 + 1, count-i2j(i, i2)