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Proof step 2 from proof abs-add-le-zero (previous step)

⊦ 0 < abs(i) + abs(j), i = 0 ∧ j = 0

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 3:
i < 0 ⊦ 0 < abs(i) + abs(j), i = 0 ∧ j = 0

Rule argument

fma: i < 0

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%lf-08: ⊦ ¬ i < j ∧ ¬ i0 < i → ¬ i0 < j; (forward) from specification int-basic1%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2 Used formulas: i < 0