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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j * j = i * j0 * j0, j < i, j < 0, 0 < i, j + 1 < i ⊦ j = 0, ∃ i0. i = i0 * i0

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%square-zero: ⊦ i * i = 0 ↔ i = 0; from specification int-mult%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1