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Proof step 6 from proof square-neq-any-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j * j = i * j0 * j0, 0 < i, j < 0 ⊦ j = 0, ∃ i0. i = i0 * i0

Rule argument

Additional information

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