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Proof step 7 from proof eq2 (previous step)

i < 0, 1 / 2 ^ abs(i) = 2, i ≤ 0 ⊦ i = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
1 / 2 ^ ~ i = 2, i < 0 ⊦ i = 1

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs