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Proof step 1 from proof count-i2j-ordered (conclusion)

⊦ ordered≤(count-i2j(i, j))

User applied rule induction

And obtained the following new goals

Proof step 2:
true, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → true → ordered≤(count-i2j(i0, j0)) ⊦ ordered≤(count-i2j(i, j))

Rule argument

indvar: i→n(j - i) wfpred: <

Additional information

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