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Proof step 17 from proof isSquare-mult (previous step)

0 ≤ i * i ∧ 0 < j0 * j0 ∧ j < 0 → (i * i) * j ≠ j0 * j0, j1 * j1 = abs(j), j0 * j0 = j * i * i, abs(j0) = abs(j1) * abs(i) ⊦ i = 0, ∃ j0. j0 * j0 = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%square-gez: ⊦ 0 ≤ i * i; from specification int-mult