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Proof step 4 from proof div2-odd-01 (previous step)

i < 0 ⊦ i + (~ i * 2 + 1) / 2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div2-odd seq: i % 2 = 0, 0 ≤ i ⊦ (i + 1) / 2 = i / 2 subst: [i] → [~ i * 2] flags: rewrite, current

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%mod-two-02: ⊦ (i * 2) % 2 = 0;%mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%mod-two-02: ⊦ (i * 2) % 2 = 0;%neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%div-mult-num-01: j % j0 = 0, j ≠ 0, j0 ≠ 0 ⊦ (i * (* j0)) / (* j) = i / j / j0;%one: ⊦ i / 1 = i;%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%neg-03: ⊦ ~ i + i = 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%zero-le: ⊦ 0 < i → (i * j ≤ 0 ↔ j ≤ 0); from specification int-mult%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: ~ i * 2