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Proof step 4 from proof shiftright-pos-div (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 ≤ i ∧ 0 < j → i >> j = i / 2 >> (j - 1)
Proof step 6:
0 ≤ i ∧ 0 < j, 0 ≤ j, 0 ≤ i, ∀ j, i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j → i0 >> j = i0 / 2 ^ j ⊦ i / 2 >> (j - 1) = i / 2 ^ j

Rule argument

name: shiftright-rec-pos rotate: F subst: [j, i] → [j, i]

Additional information

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