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Proof step 34 from proof congmod-def-mod-02 (
previous step
)
0 ≤ i % abs(j), i0 < 0, i < 0, 0 ≤ i % j, 0 ≤ i0 % j ⊦ j = 0, i % abs(j) = i0 % j ↔ i % j = i0 % j
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 35
:
⊦ 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
Additional information
Used terms: j, i