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Proof step 11 from proof div-nearer-zero (previous step)

0 < i / (0 - j) * (0 - j), i < i / (0 - j) * (0 - j), j < 0, 0 ≤ i ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
i / j < 0, i < j * i / j, j < 0, 0 ≤ i ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i < 0 → (0 < i * j ↔ 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)% ⊦ 0 - i = ~ i; (s)