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Proof step 3 from proof abs-succ (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: j < 0

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j;%mone-06: ⊦ ~1 < i ↔ 0 ≤ i;%lels-03: ⊦ i < j → (j ≤ i ↔ false);%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss) Used formulas: j < 0