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Proof step 2 from proof square-neq-any-pos (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
0 < i, 0 < j, 0 < j0, j * j = i * j0 * j0 ⊦ ∃ i0. i = i0 * i0
Proof step 4:
j * j = i * j0 * j0, 0 < i, j ≠ 0 ∨ j0 ≠ 0 ⊦ ∃ i0. i = i0 * i0, 0 < j ∧ 0 < j0

Rule argument

spec: <toplevel> 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]

Additional information

Used terms: i, j, j0