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Proof step 1 from proof mod-pred-01 (conclusion)

⊦ (i - 1) % j + 1 = j ∧ 0 < i ∧ 0 < j → i % j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
(i - 1) % j + 1 = j, 0 < j, 0 < i, 0 ≤ (i - 1) % j ⊦ i % j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1