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Proof step 3 from proof count-i2j-max (previous step)

i ≤ j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → max(count-i2j(i0, j0)) = j0 ⊦ max(count-i2j(i, j)) = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i ≤ j → count-i2j(i, j) = i ' + count-i2j(i + 1, j)
Proof step 5:
i ≤ j, i ≤ j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → max(count-i2j(i0, j0)) = j0 ⊦ max(i ' + count-i2j(i + 1, j)) = j

Rule argument

spec: inst: name: count-rec seq: ⊦ i ≤ j → count-i2j(i, j) = i ' + count-i2j(i + 1, j) subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used terms: j, i