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Proof step 3 from proof count-i2j-no-dups (previous step)

dups(count-i2j(i, j)), ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → dups(count-i2j(i0, j0)) → false ⊦ false

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information