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

¬ ¬ i < max(count-i2j(i + 1, j)), i < j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → max(count-i2j(i0, j0)) = j0 ⊦ max(i, max(count-i2j(i + 1, j))) = j, i = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i < max(count-i2j(i + 1, j)), i < j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → max(count-i2j(i0, j0)) = j0 ⊦ i = j, max(count-i2j(i + 1, j)) = j

Rule argument

Additional information

Used simplifier rules: max-c: ⊦ max(i, j) = max(j, i); from specification int-min%max-a: ⊦ max(i, max(j, j0)) = max(max(i, j), j0); from specification int-min%Max-right: ⊦ j < i → max(j, i) = i; from specification int-min