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Proof step 21 from proof congmod-def-mod-02 (previous step)

0 ≤ i0 % abs(j), 0 ≤ i % abs(j), i0 < 0, i < 0, 0 ≤ i % j ⊦ j = 0, i % abs(j) = i0 |%| abs(j) ↔ i % j = i0 |%| j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ j ≠ 0 → i % abs(j) = i % j
Proof step 23:
j ≠ 0, 0 ≤ i0 % j, 0 ≤ i % abs(j), i0 < 0, i < 0, 0 ≤ i % j ⊦ j = 0, i % abs(j) = i0 |%| abs(j) ↔ i % j = i0 |%| j

Rule argument

spec: int-div inst: name: mod-abs seq: ⊦ j ≠ 0 → i % abs(j) = i % j subst: [j, i] → [j, i0] flags: rewrite

Additional information

Used terms: j, i0