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Proof step 5 from proof shiftright-neg-div (previous step)

i + 2 ^ (abs(j) - 1) < 0, ∀ i, j0. i→n(j0) < i→n(j) → i + 2 ^ (abs(j0) - 1) < 0 → j0 = 0 ∨ i >> abs(j0) = (i + 1 - 2 ^ abs(j0)) / 2 ^ abs(j0) ⊦ j = 0 ∨ i >> abs(j) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i < 0 ∧ j ≠ 0 → i >> j = (i - 1) / 2 >> (abs(j) - 1)
Proof step 7:
i < 0 ∧ abs(j) ≠ 0, i + 2 ^ (abs(j) - 1) < 0, ∀ i, j0. i→n(j0) < i→n(j) → i + 2 ^ (abs(j0) - 1) < 0 → j0 = 0 ∨ i >> abs(j0) = (i + 1 - 2 ^ abs(j0)) / 2 ^ abs(j0) ⊦ j = 0 ∨ (i - 1) / 2 >> (abs(abs(j)) - 1) = (i + 1 - 2 ^ abs(j)) / 2 ^ abs(j)

Rule argument

name: shiftright-rec-neg rotate: F subst: [j, i] → [abs(j), i]

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ abs(i) = 0 ↔ i = 0;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i + j < j ↔ i < 0;% ⊦ ¬ 2 ^ i < 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: abs(j), i