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Proof step 2 from proof i2n-mod (previous step)

0 < j, 0 ≤ i ⊦ i→n(i % j) = i→n(i) % i→n(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < j, 0 ≤ i ⊦ i→n(i % j) = i→n(i) % i→n(j)

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs