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Proof step 3 from proof mod-abs (previous step)

⊦ i % abs(j) = i % j, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ¬ j < 0 → abs(j) = j

Rule argument

spec: int-abs name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [j] flags: <all paths>

Additional information

Used simplifier rules: mod-neg: j ≠ 0 ⊦ i % ~ j = i % j;% ⊦ i < 0 → abs(i) = ~ i; (s) Used terms: j