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Proof step 2 from proof abs-eqall (previous step)

i < 0 ⊦ abs(i * j) = abs(i) * abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs