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Proof step 8 from proof div2 (previous step)

i0 < 0, i0 + 1 < 0 ⊦ 1 / 2 ^ abs(i0 + 1) = ~1, 1 / 2 ^ abs(i0 + 1) = 1, 1 / 2 ^ abs(i0 + 1) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
i0 < ~1, i0 < 0 ⊦ 1 / 2 ^ ~(i0 + 1) = 0, 1 / 2 ^ ~(i0 + 1) = ~1, 2 ^ ~(i0 + 1) = 1

Rule argument

Additional information

Used simplifier rules: zero-02: ⊦ j < 0 → (i ^ ~ j = 0 ↔ i = 0);%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)%j ≠ 0 ⊦ 1 / j = 1 ↔ j = 1; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ~ i - j = ~(i + j); (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i < 0 → abs(i + 1) = abs(i) - 1; (s)