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Proof step 13 from proof square-neq-any-pos (previous step)

j * j = i * j0 * j0, 0 < i, 0 ≤ j, 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 14:
j * j = i * j0 * j0, 0 < i, 0 ≤ j, j ≠ 0 ∨ j0 ≠ 0 ⊦ 0 = j, ∃ i0. i = i0 * i0, 0 < j ∧ 0 < j0

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% ⊦ 0 < i → (i * j = 0 ↔ j = 0); (s)% ⊦ i * i = 0 ↔ i = 0; (s) Used formulas: 0 ≠ j