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Proof step 12 from proof square-neq-any-pos (previous step)

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

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 0 ≠ j

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%square-zero: ⊦ i * i = 0 ↔ i = 0; from specification int-mult Used formulas: 0 ≠ j