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Proof step 6 from proof square-ex (previous step)

0 < (j + 1) * (j + 1), 0 < j, 1 < (j + 1) * (j + 1), 0 < i0, i0 < (j + 1) * (j + 1), j * j ≤ 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 7:
0 < (j + 1) * (j + 1), 0 < j, 1 < (j + 1) * (j + 1), 0 < i0, i0 < (j + 1) * (j + 1), j * j ≤ i0, 0 ≤ i0 ⊦ 0 < j ∧ j * j ≤ i0 + 1 ∧ i0 + 1 < (j + 1) * (j + 1), ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1)

Rule argument

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

Additional information

Used terms: j