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Proof step 5 from proof two-minus (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i < 0, 2 ^ abs(i) ≤ 1 ⊦ 1 / 2 ^ abs(i) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 1 < j → 1 / j = 0; (s)