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Proof step 11 from proof square-neq-any (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

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