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Proof step 10 from proof shiftright-pos-div (previous step)

0 ≤ j, 0 ≤ i, 0 < j, ∀ j, i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j → i0 >> j = i0 / 2 ^ j ⊦ (i / 2) / 2 ^ (j - 1) = i / 2 ^ j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ 0 ≤ i ∧ 0 ≤ j → (j / 2) / 2 ^ i = j / 2 ^ (i + 1) subst: [i, j] → [j - 1, i] rotate: F spec: int-pot inst: name: div2-2pot-01

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i) = j;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i / 1 = i; Used terms: j - 1, i