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Proof step 22 from proof equal-minus-one (previous step)

abs(i ^ i0) = 1, i + i ^ i0 = 0, 0 ≤ i0 ⊦ i0 = 0, i = 0, i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
abs(i ^ i0) = 1, i + i ^ i0 = 0, 0 < i0 ⊦ i0 = 0, i = 0, i = ~1

Rule argument

Additional information

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