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Proof step 3 from proof square-neq-any (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
j * j = i * j0 * j0, i ≤ 0, j ≠ 0 ∨ j0 ≠ 0 ⊦ ∃ i0. i = i0 * i0

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs