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Proof step 3 from proof absone-01 (previous step)

abs(i) = 2 + j, j < 0 ⊦ i = 0, j = ~1

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
abs(i) = 2 + j, j < 0, i < 0 ⊦ i = 0, j = ~1

Rule argument

fma: i < 0

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ i < 0 → (~2 < i ↔ i = ~1); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: i < 0