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Proof step 14 from proof modmod-mult-07 (previous step)

¬(0 < j ∧ 0 ≤ ~ i ∧ 0 ≤ i0), j ≠ 0, j ≠ 0 ∧ i < 0, i < j, i < 0, 0 < j, i + 1 < j ⊦ ~((~ i % j * i0) % j) = (i * i0) % j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-neg-01: 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-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; 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%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%neg-02: ⊦ ~ ~ i = i; from specification int-abs%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