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Proof step 6 from proof square-neq-any-pos (previous step)

j * j = i * j0 * j0, j < i, j < 0, 0 < i, j + 1 < i ⊦ j = 0, ∃ i0. i = i0 * i0

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
0 < i, 0 < j, 0 < j0, j * j = i * j0 * j0 ⊦ ∃ i0. i = i0 * i0
Proof step 8:
j * j = i * j0 * j0, j < i, j < 0, 0 < i, j + 1 < i ⊦ 0 < ~ j ∧ 0 < j0 ∧ ~ j * ~ j = i * j0 * j0, j = 0, ∃ 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