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

abs(i) + abs(j) = 0 ⊦ i = 0 ∧ 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, i = 0 ∧ j = 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-eq: ⊦ i - j = i0 ↔ i = i0 + j;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0;%absls: ⊦ ¬ abs(i) < 0;%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