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Proof step 3 from proof absls-02 (previous step)

⊦ abs(i) < abs(j) ↔ j < i, 0 < j, 0 < i

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: i < 0

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j;%abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ ¬ i < j → (j < i ↔ i ≠ j); (ss)% ⊦ 0 + i = i; (s)%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