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Proof step 2 from proof abs-add-pos (previous step)

abs(i) + abs(j) < 0 ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: i < 0

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%sub-zero: ⊦ i + (0 - i0) = i - i0;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%abs-11: ⊦ j < 0 → ¬ abs(i) < j;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2 Used formulas: i < 0