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Proof step 6 from proof pot-root-est (previous step)

0 < i, 1024 ≤ i, 4 ^ (i / 3) ≤ (2 * i) ^ sqrt(2 * i), 0 ≤ i, ∃ j, j0. 2 * i = j * j + j0 ∧ j0 ≤ 2 * j ∧ 0 ≤ j ∧ 0 ≤ j0 ∧ j = sqrt(2 * i) ⊦

User applied rule exists left

And obtained the following new goals

Proof step 7:
2 * i = j * j + j0, j = sqrt(2 * i), 0 < i, 0 ≤ i, 4 ^ (i / 3) ≤ (2 * i) ^ sqrt(2 * i), 1024 ≤ i, j0 ≤ 2 * j, 0 ≤ j, 0 ≤ j0 ⊦

Rule argument

pos: left 5

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