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Proof step 6 from proof div-twoodd-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);%neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ ~ ~ i = i; (s)% ⊦ ~ i * j = ~(i * j); (s)% ⊦ i + ~ j = i - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ 2 * i = 0 ↔ i = 0; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)% ⊦ i + 0 = i; (s) Used terms: j, ~ i