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Proof step 6 from proof div-abs (previous step)

⊦ i = abs(i), j = 0, ∃ i0. i * i0 = abs(i) * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i < 0 ⊦ j = 0, ∃ i0. i * j + i * i0 = 0

Rule argument

Additional information

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