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Proof step 35 from proof congmod-def-mod-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

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 simplifier rules: mod-zero-06: ⊦ 0 ≤ i ∧ j ≠ 0 → ¬ i % j < 0; from specification int-div Used terms: j, i