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Proof step 2 from proof mod-abs-intro (
previous step
)
⊦ i % j = i % abs(j), j = 0
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 3
:
⊦ j ≠ 0 → i % abs(j) = i % j
Rule argument
spec: int-div inst: name: mod-abs seq: ⊦ j ≠ 0 → i % abs(j) = i % j subst: [j, i] → [j, i] flags: rewrite, current
Additional information
Used terms: j, i