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

i < 0, 0 < j ⊦ j = 0, ~((~ i % j * i0) % j) = (i * i0) % j, 0 < j ∧ 0 ≤ ~ i ∧ 0 ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i < 0, 0 < j, i0 < 0, i ≤ 0 ⊦ (i % j * i0) % j = (i * i0) % j, j = 0

Rule argument

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)% ⊦ ~ ~ i = i; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ ~ i * j = ~(i * j); (s)