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Proof step 3 from proof div-neg (previous step)

⊦ (∃ i0. i * i0 = ~ j) ↔ i | j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∃ i0. j + i * i0 = 0) ↔ i | j

Rule argument

Additional information

Used simplifier rules: negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs