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Proof step 10 from proof modmod-mult-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 < j0, j < 0, i < 0 ⊦ (i % (j * j0)) % j = i % j, j = 0

Rule argument

Additional information

Used simplifier rules: mod-neg: j ≠ 0 ⊦ i % ~ j = i % j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i * 0 = 0; (s)% ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; (ws)%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)% ⊦ 0 < ~ i ↔ i < 0; (s)