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Proof step 4 from proof mod-i2n (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i, 0 < j ⊦ i→n(i % j) = i→n(i) % i→n(j)

Rule argument

Additional information

Used simplifier rules: mod-zero: ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j; from specification int-div