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Proof step 2 from proof absone-01 (previous step)

abs(i) = j + 1 + 1, j < 0 ⊦ i = 0, j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
abs(i) = 2 + j, j < 0 ⊦ i = 0, j = ~1

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2