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Proof step 10 from proof square-neq-any-pos (previous step)

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

User applied rule apply lemma

And obtained the following new goals

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 simplifier rules: ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ ~ i * j = ~(i * j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ ~ ~ i = i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i * 0 = 0; (s)% ⊦ i * i = 0 ↔ i = 0; (s) Used terms: i, ~ j, ~ j0