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Proof step 3 from proof mod-zero-09 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ abs(i) < j → i % j = i

Rule argument

spec: <toplevel> name: mod-zero-05 seq: ⊦ abs(i) < j → i % j = i subst: [j, i] → [j, i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: j, i