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Proof step 5 from proof pot-root-est (previous step)

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

User applied rule exists left

And obtained the following new goals

Proof step 6:
2 * i = j * j + j0 ∧ j0 ≤ 2 * j ∧ 0 ≤ j ∧ 0 ≤ j0 ∧ j = sqrt(2 * i), 4 ^ (i / 3) ≤ (2 * i) ^ sqrt(2 * i), 1024 ≤ i, 0 ≤ i, 0 < i ⊦

Rule argument

pos: left 1

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