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Proof step 19 from proof divmod2-2pot-neg (previous step)

j0 < i, j0 ≤ 0, i0 * 2 ^ i + j0 < 0, abs(j0) < 2 ^ i, 1 < i, 0 < i, 0 ≤ i, i0 < 0 ⊦ j0 = 0, ((j0 - 1) / 2) % 2 = ((j0 - 1) / 2 + i0 * 2 ^ (i - 1)) % 2, i0 = 0, (i0 * 2 ^ (i - 1)) % 2 = 0 ∧ j0 < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
1 < i, 0 < i, 0 ≤ i, i0 < 0, j0 < i, i0 * 2 ^ i + j0 < 0, j0 < 0, 0 < 2 ^ i + j0 ⊦ j0 = 0, ((j0 - 1) / 2) % 2 = ((j0 - 1) / 2 + i0 * 2 ^ (i - 1)) % 2, i0 = 0, (i0 * 2 ^ (i - 1)) % 2 = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs