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Proof step 6 from proof div-twoodd-05 (previous step)

i < 0, 0 < j ⊦ (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

Proof step 7:
0 < j ⊦ 1 / (2 * j) = 0

Rule argument

Additional information

Used simplifier rules: num-neg: j ≠ 0 ⊦ i < 0 → (* i) / j = ~(~ i / j);%zero: j ≠ 0 ⊦ 0 / j = 0;% ⊦ ~ i = 0 ↔ i = 0; (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%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%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)%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)