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Proof step 8 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(i, max(count-i2j(i + 1, j))) = j, i = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ¬ j < i → max(j, i) = j
Proof step 10:
¬ ¬ 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
Proof step 12:
¬ 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, i = j

Rule argument

spec: int-min inst: name: Max-left seq: ⊦ ¬ j < i → max(j, i) = j subst: [i, j] → [max(count-i2j(i + 1, j)), i] flags: rewrite

Additional information

Used terms: max(count-i2j(i + 1, j)), i