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

abs(i) = 2 + j, j < 0 ⊦ j = ~1, i = 0

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

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;%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%minus-one: ⊦ i < 0 → (~2 < i ↔ i = ~1); from specification int-basic1%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