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Proof step 13 from proof count-i2j-no-dups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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