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Proof step 10 from proof count-i2j-ordered (previous step)

i ≤ i1 + i, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → true → ordered≤(count-i2j(i0, j0)) ⊦ ordered≤(i ' + count-i2j(i + 1, i1 + i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs