| ⊦ i < 0 ∧ 0 < j ∧ j < 32 → i i>> j = (i - (2 ^ j - 1)) / 2 ^ j |
true, ∀ i, j0. i→n(j0) < i→n(j) → true → i < 0 ∧ 0 < j0 ∧ j0 < 32 → i i>> j0 = (i - (2 ^ j0 - 1)) / 2 ^ j0 ⊦ i < 0 ∧ 0 < j ∧ j < 32 → i i>> j = (i - (2 ^ j - 1)) / 2 ^ j
Used simplifier rules: <; (wellfounded) Used terms: i→n(j)