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Proof step 5 from proof square-neq-any (previous step)

j * j = i * j0 * j0, i ≤ 0, j ≠ 0 ∨ j0 ≠ 0 ⊦ i = 0, ∃ i0. i = i0 * i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j * j = i * j0 * j0, i < 0, j ≠ 0 ∨ j0 ≠ 0 ⊦ i = 0, ∃ i0. i = i0 * i0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs