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Proof step 1 from proof square-neq-any-pos (conclusion)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
0 < i, 0 < j, 0 < j0, j * j = i * j0 * j0 ⊦ ∃ i0. i = i0 * i0
Proof step 3:
0 < i, j ≠ 0 ∨ j0 ≠ 0, j * j = i * j0 * j0 ⊦ 0 < j ∧ 0 < j0, ∃ i0. i = i0 * i0

Rule argument

spec: int-gcd inst: name: square-neq-any-pos-01 seq: 0 < i, 0 < j, 0 < j0, j * j = i * j0 * j0 ⊦ ∃ i0. i = i0 * i0 subst: [i, j, j0] → [i, j, j0] flags: current, []

Additional information

Used terms: i, j, j0