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Proof step 1 from proof mod-abs-pos (conclusion)

⊦ 0 ≤ i ∧ j ≠ 0 → i |%| abs(j) = i |%| j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ i |%| abs(j) = i |%| j, j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs