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Proof step 1 from proof shiftright-neg-div-02 (conclusion)

⊦ i % 2 ^ abs(j) ≠ 0 ∧ i < ~(2 ^ (abs(j) - 1)) ∧ j ≠ 0 → i >> j = i / 2 ^ abs(j) - 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ i < ~(2 ^ (abs(j) - 1)) ∧ j ≠ 0 → i >> j = (i - (2 ^ abs(j) - 1)) / 2 ^ abs(j)
Proof step 3:
i < ~(2 ^ (abs(j) - 1)) ∧ j ≠ 0 ⊦ i % 2 ^ abs(j) ≠ 0 ∧ i < ~(2 ^ (abs(j) - 1)) ∧ j ≠ 0 → (i - (2 ^ abs(j) - 1)) / 2 ^ abs(j) = i / 2 ^ abs(j) - 1

Rule argument

name: shiftright-neg-div rotate: F subst: [j, i] → [j, i]

Additional information

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