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Proof step 8 from proof square-ex (previous step)

(j + 1) * (j + 1) = i0 + 1, 0 < j, j * j ≤ i0, 0 ≤ i0, 0 < i0 ⊦ ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1)

User applied rule exists right

And obtained the following new goals

Proof step 9:
(j + 1) * (j + 1) = i0 + 1, 0 < j, j * j ≤ i0, 0 ≤ i0, 0 < i0 ⊦ 0 < j + 1 ∧ (j + 1) * (j + 1) ≤ i0 + 1 ∧ i0 + 1 < ((j + 1) + 1) * ((j + 1) + 1)

Rule argument

pos: right 1 terms: j + 1 abort: F discard: T computed: F

Additional information

Used terms: j + 1