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Proof step 2 from proof mod-abs-pos (previous step)

0 ≤ i ⊦ i |%| abs(j) = i |%| j, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ j < 0 → abs(j) = j
Proof step 4:
¬ ¬ j < 0, 0 ≤ i ⊦ i |%| abs(j) = i |%| j, j = 0

Rule argument

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

Additional information

Used terms: j