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Proof step 15 from proof pot-root-est (previous step)

2 * i = j * j + j0, j = sqrt(2 * i), (2 * i) ^ j < ((j + 1) ^ 2) ^ j, 0 < i, 2 * i < (j + 1) ^ 2, (j ^ 2 / 2) / 3 ≤ i / 3, j ^ 2 / 2 ≤ i, 0 ≤ j0, 0 ≤ j, j0 ≤ 2 * j, 1024 ≤ i, 4 ^ (i / 3) ≤ (2 * i) ^ sqrt(2 * i), 0 ≤ i, j ^ 2 ≤ 2 * i, j * j / 6 ≤ (j ^ 2 / 2) / 3 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 16:
2 * i = j * j + j0, j = sqrt(2 * i), 2 * i < (j + 1) ^ 2, 0 < i, (2 * i) ^ j < ((j + 1) ^ 2) ^ j, j * j / 6 ≤ (j ^ 2 / 2) / 3, j ^ 2 ≤ 2 * i, 0 ≤ i, 4 ^ (i / 3) ≤ (2 * i) ^ sqrt(2 * i), 1024 ≤ i, j0 ≤ 2 * j, 0 ≤ j, 0 ≤ j0, j ^ 2 / 2 ≤ i, (j ^ 2 / 2) / 3 ≤ i / 3, 4 ^ (j * j / 6) ≤ 4 ^ (i / 3) ⊦
Proof step 41:
2 * i = j * j + j0, j = sqrt(2 * i), 2 * i < (j + 1) ^ 2, 0 < i, (2 * i) ^ j < ((j + 1) ^ 2) ^ j, j * j / 6 ≤ (j ^ 2 / 2) / 3, j ^ 2 ≤ 2 * i, 0 ≤ i, 4 ^ (i / 3) ≤ (2 * i) ^ sqrt(2 * i), 1024 ≤ i, j0 ≤ 2 * j, 0 ≤ j, 0 ≤ j0, j ^ 2 / 2 ≤ i, (j ^ 2 / 2) / 3 ≤ i / 3 ⊦ 4 ^ (j * j / 6) ≤ 4 ^ (i / 3)

Rule argument

fma: 4 ^ (j * j / 6) ≤ 4 ^ (i / 3)

Additional information

Used formulas: 4 ^ (j * j / 6) ≤ 4 ^ (i / 3)