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Proof step 2 from proof mod-pred-02 (previous step)

0 < i, 0 < j ⊦ (i - 1) % j = i % j - 1, i % j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < j, 0 < i ⊦ i % j = 0, i % j = (i - 1) % j + 1

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i - j = i0 ↔ i = i0 + j; (s)