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Proof step 29 from proof intshiftrsign-2pot-neg-01 (previous step)

0 < j, abs(j1) < 2, i0 < j, i0 ≤ 0, j1 < j, j1 ≤ 0, j < 32, i0 * 2 + j1 < ~1, i0 * 2 + j1 < 0, ∀ i, j0. i→n(j0) < i→n(j) → true → i < 0 ∧ 0 < j0 ∧ j0 < 32 → i i>> j0 = (i - (2 ^ j0 - 1)) / 2 ^ j0 ⊦ j1 = ~1, j = 1, (i0 + 1 - 2 ^ (j - 1)) / 2 ^ (j - 1) = (2 * (i0 + 1) + j1 - 2 ^ j) / 2 ^ j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ j < 0 → abs(j) = 0 - j
Proof step 31:
¬ j1 < 0, 0 < j, abs(j1) < 2, i0 < j, i0 ≤ 0, j1 < j, j1 ≤ 0, j < 32, i0 * 2 + j1 < ~1, i0 * 2 + j1 < 0, ∀ i, j0. i→n(j0) < i→n(j) → true → i < 0 ∧ 0 < j0 ∧ j0 < 32 → i i>> j0 = (i - (2 ^ j0 - 1)) / 2 ^ j0 ⊦ j1 = ~1, j = 1, (i0 + 1 - 2 ^ (j - 1)) / 2 ^ (j - 1) = (2 * (i0 + 1) + j1 - 2 ^ j) / 2 ^ j

Rule argument

seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j1] rotate: F spec: int-abs inst: name: Abs-neg

Additional information

Used simplifier rules: ⊦ 0 - i = ~ i;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i < 0 → (~2 < i ↔ i = ~1);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: j1