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Proof step 8 from proof count-i2j-min (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ¬ j < i → min(j, i) = i
Proof step 10:
¬ i < min(count-i2j(i + 1, j)), i < j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → min(count-i2j(i0, j0)) = i0 ⊦ min(count-i2j(i + 1, j)) = i, i = j

Rule argument

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

Additional information

Used simplifier rules: Min-left: ⊦ j < i → min(j, i) = j; from specification int-min%min-a: ⊦ min(i, min(j, j0)) = min(min(i, j), j0); from specification int-min%min-c: ⊦ min(i, j) = min(j, i); from specification int-min Used terms: min(count-i2j(i + 1, j)), i