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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs