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

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

User applied rule case distinction

And obtained the following new goals

Proof step 6:
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) ⊦

Rule argument

pos: left 5

Additional information

Used simplifier rules: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult