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Proof step 15 from proof isSquare-mult (previous step)

j1 * j1 = abs(j), j0 * j0 = j * i * i, abs(j0) = abs(j1) * abs(i) ⊦ i = 0, ∃ j0. j0 * j0 = j

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 16:
⊦ 0 ≤ i ∧ 0 < j ∧ i1 < 0 → i * i1 ≠ j
Proof step 17:
0 ≤ i * i ∧ 0 < j0 * j0 ∧ j < 0 → (i * i) * j ≠ j0 * j0, j1 * j1 = abs(j), j0 * j0 = j * i * i, abs(j0) = abs(j1) * abs(i) ⊦ i = 0, ∃ j0. j0 * j0 = j

Rule argument

spec: int-mult inst: name: mult-pos seq: ⊦ 0 ≤ i ∧ 0 < j ∧ i1 < 0 → i * i1 ≠ j subst: [i, j, i1] → [i * i, j0 * j0, j] flags: []

Additional information

Used terms: i * i, j0 * j0, j