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Proof step 3 from proof absa (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: i + j < 0

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%lf-08: ⊦ ¬ i < j ∧ ¬ i0 < i → ¬ i0 < j; (forward) from specification int-basic1% ⊦ i + j < j ↔ i < 0; (s)% ⊦ ¬ i < j → (j < i ↔ i ≠ j); (ss)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lf-06: ⊦ ¬ j < i ∧ i = i0 → ¬ j < i0; (forward) from specification int-basic1%lf-05: ⊦ ¬ j < i ∧ j = i0 → ¬ i0 < i; (forward) from specification int-basic1 Used formulas: i + j < 0