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Proof step 13 from proof divide-list-at-value (previous step)

j < i, perm(ints0, ints1 + ints), min(i, min(ints)) ≤ j, j < min(ints), min(i, min(ints)) < i, min(i, min(ints)) < min(ints), ints1 = [] ∨ max(ints1) ≤ j ⊦ ints = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ ¬ j < i → min(j, i) = i

Rule argument

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

Additional information

Used simplifier rules: lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%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(ints), i