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Proof step 10 from proof mod2-2pot-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
0 ≤ i, 0 < j ⊦ (2 * i + 1) % (2 * j) = i % j * 2 + 1
Proof step 12:
0 < 2 ^ i, 0 ≤ i, 0 ≤ i0 ⊦ (i0 % 2 ^ i * 2 + 1) / 2 = i0 % 2 ^ i

Rule argument

spec: int-div name: mod-twoodd seq: 0 ≤ i, 0 < j ⊦ (2 * i + 1) % (2 * j) = i % j * 2 + 1 subst: [j, i] → [2 ^ i, i0] flags: <all paths>

Additional information

Used simplifier rules: two-lez: ⊦ 0 ≤ i → ¬ 2 ^ i ≤ 0;% ⊦ i < j ↔ ¬ j ≤ i; (ss)%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2 ^ i, i0