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Proof step 16 from proof count-i2j-no-dups (previous step)

¬ ¬ 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 ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: count-i2j-in: ⊦ i0 ∈ count-i2j(i, j) ↔ i ≤ i0 ∧ i0 ≤ j;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs