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Proof step 13 from proof div-twoodd-07 (previous step)

¬(i + 1 < 0 ∧ 0 < ~ j), j * 2 < 0, i < 0, j < 0 ⊦ ~((2 * (i + 1) - 1) / (2 * ~ j)) = (i + 1) / 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;%neg-10: ⊦ ~ i = 0 ↔ i = 0; from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%mone-08: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; 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%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%neg-02: ⊦ ~ ~ i = i; from specification int-abs%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%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