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Proof step 6 from proof modmod-mult-03 (previous step)

j < 0, 0 < j0 ⊦ (i % (j * j0)) % j = i % j, j = 0, 0 < j ∧ 0 ≤ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j < j0, j < 0, 0 < j0, j + 1 < j0 ⊦ j = 0, (i % (j * j0)) % j = i % j

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1