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Proof step 2 from proof mod-zero-09 (previous step)

i < j, 0 < i, 0 < j, 1 < j ⊦ i % j = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ abs(i) < j → i % j = i

Rule argument

spec: inst: name: mod-zero-05 seq: ⊦ abs(i) < j → i % j = i subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used simplifier rules: abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: j, i