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Proof step 9 from proof modmod-mult-03 (previous step)

0 < ~ j ∧ 0 ≤ i ∧ 0 < j0 → (i % (~ j * j0)) % ~ j = i % ~ j, j < j0, j < 0, 0 < j0, j + 1 < j0 ⊦ j = 0, (i % (j * j0)) % j = i % j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i < j0, i < 0, j + 1 < j0, 0 < j0, j < 0, j < j0, i + 1 < j0 ⊦ (i % (j * j0)) % j = i % j, j = 0

Rule argument

Additional information

Used simplifier rules: mod-neg: j ≠ 0 ⊦ i % ~ j = i % j;%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%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs