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Proof step 6 from proof div-twoodd-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: div2-odd seq: i % 2 = 0, 0 ≤ i ⊦ (i + 1) / 2 = i / 2 subst: [i] → [~ i] flags: <all paths>

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%Div-left: ⊦ j ≠ 0 → ~ i / j = ~(i / j);%two-even: ⊦ i % 2 = 0 → (i / 2 = j ↔ i = 2 * j);%divmult: j ≠ 0 ⊦ i % j = 0 → i / j * j = i;% ⊦ ~ i = 0 ↔ i = 0; (s)% ⊦ ~ ~ i = i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ 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)% ⊦ 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: ~ i