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Proof step 7 from proof div2-2pot-03 (previous step)

0 ≤ j, 0 ≤ i0 ⊦ (j * 2 ^ i0 * 2 + 1) / 2 = j * 2 ^ i0

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] → [2 * 2 ^ i0 * j] flags: rewrite

Additional information

Used simplifier rules: two-lsz: ⊦ ¬ 2 ^ i < 0;%mod-two-02: ⊦ (i * 2) % 2 = 0; from specification int-div%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i; from specification int-div%zero-ge-02: ⊦ 0 ≤ i → (0 ≤ i * j ↔ ¬ ¬(i = 0 ∨ 0 ≤ j)); from specification int-mult%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2 * 2 ^ i0 * j