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Proof step 5 from proof absa (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
i + j < 0, i < 0 ⊦ 0 = (abs(i) + abs(j)) + i + j, 0 < i, 0 < j

Rule argument

fma: i < 0

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%Abs-neg: ⊦ j < 0 → abs(j) = 0 - j;%sub-add-01: ⊦ i + (j - i0) = i + j - i0;%sub-sub-03: ⊦ i - i = 0;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ¬ i < j → (j < i ↔ i ≠ j); (ss)%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%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%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 + i = i; (s) Used formulas: i < 0