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Proof step 14 from proof div-twoodd-07 (previous step)

i < 0, j < 0, j * 2 < 0 ⊦ ~((2 * (i + 1) - 1) / (2 * ~ j)) = (i + 1) / j, i + 1 < 0 ∧ 0 < ~ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);%num-neg: j ≠ 0 ⊦ i < 0 → (* i) / j = ~(~ i / j);%zero: j ≠ 0 ⊦ 0 / j = 0;% ⊦ ~ i = 0 ↔ i = 0; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < ~ i ↔ i < 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)% ⊦ ~ ~ i = i; (s)% ⊦ 2 * i = 0 ↔ i = 0; (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)