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Proof step 2 from proof mod-zero-05 (previous step)

abs(i) < j ⊦ i % j = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
0 < j, 0 ≤ i, i < j ⊦ i % j = i
Proof step 4:
¬(0 < j ∧ 0 ≤ i ∧ i < j), abs(i) < j ⊦ i % j = i

Rule argument

spec: int-div inst: name: mod-zero-03 seq: 0 < j, 0 ≤ i, i < j ⊦ i % j = i subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used terms: j, i