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Proof step 2 from proof div-twoodd-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
j ≠ 0 ⊦ i < 0 → (* i) / j = ~(~ i / j)
Proof step 4:
i < 0, 0 < j, 2 * i - 1 < 0 ⊦ 2 * j = 0, ~(~(2 * i - 1) / (2 * j)) = i / j

Rule argument

spec: <toplevel> name: num-neg seq: j ≠ 0 ⊦ i < 0 → (* i) / j = ~(~ i / j) subst: [j, i] → [2 * j, 2 * i - 1] flags: <all paths>

Additional information

Used simplifier rules: nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);% ⊦ 2 * i = 0 ↔ i = 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i - j < i0 ↔ i < j + i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: 2 * j, 2 * i - 1