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Proof step 2 from proof count-i2j-no-dups (previous step)

dups(count-i2j(i, j)) ⊦

User applied rule induction

And obtained the following new goals

Proof step 3:
dups(count-i2j(i, j)), ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → dups(count-i2j(i0, j0)) → false ⊦ false

Rule argument

indvar: i→n(j - i) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j - i)