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Proof step 3 from proof modmod-mult-07 (previous step)

¬(0 < j ∧ 0 ≤ i ∧ 0 ≤ i0), 0 < j ⊦ (i % j * i0) % j = (i * i0) % j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 < j ⊦ (i % j * i0) % j = (i * i0) % j, 0 ≤ i ∧ 0 ≤ i0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs