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

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

User applied rule cut formula

And obtained the following new goals

Proof step 6:
j * j = i * j0 * j0, 0 < i, j < 0, j ≠ 0 ∨ j0 ≠ 0 ⊦ ∃ i0. i = i0 * i0, 0 < j ∧ 0 < j0
Proof step 12:
j * j = i * j0 * j0, 0 < i, j ≠ 0 ∨ j0 ≠ 0 ⊦ j < 0, ∃ i0. i = i0 * i0, 0 < j ∧ 0 < j0

Rule argument

fma: j < 0

Additional information

Used formulas: j < 0