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Proof step 10 from proof root-elim (previous step)

i < (sqrt(i) + 1) * (sqrt(i) + 1), sqrt(i) * sqrt(i) ≤ i, 0 ≤ i, 0 < (sqrt(i) + 1) * (sqrt(i) + 1), sqrt(i) < abs(sqrt(i) + 1) ⊦ ∃ j0. i = sqrt(i) * sqrt(i) + j0 ∧ j0 ≤ 2 * sqrt(i) ∧ 0 ≤ sqrt(i) ∧ 0 ≤ j0

User applied rule cut formula

And obtained the following new goals

Proof step 11:
(sqrt(i) + 1) * (sqrt(i) + 1) = sqrt(i) * sqrt(i) + 2 * sqrt(i) + 1, i < (sqrt(i) + 1) * (sqrt(i) + 1), sqrt(i) * sqrt(i) ≤ i, 0 ≤ i, 0 < (sqrt(i) + 1) * (sqrt(i) + 1), sqrt(i) < abs(sqrt(i) + 1) ⊦ ∃ j0. i = sqrt(i) * sqrt(i) + j0 ∧ j0 ≤ 2 * sqrt(i) ∧ 0 ≤ sqrt(i) ∧ 0 ≤ j0
Proof step 14:
i < (sqrt(i) + 1) * (sqrt(i) + 1), sqrt(i) * sqrt(i) ≤ i, 0 ≤ i, 0 < (sqrt(i) + 1) * (sqrt(i) + 1), sqrt(i) < abs(sqrt(i) + 1) ⊦ (sqrt(i) + 1) * (sqrt(i) + 1) = sqrt(i) * sqrt(i) + 2 * sqrt(i) + 1, ∃ j0. i = sqrt(i) * sqrt(i) + j0 ∧ j0 ≤ 2 * sqrt(i) ∧ 0 ≤ sqrt(i) ∧ 0 ≤ j0

Rule argument

fma: (sqrt(i) + 1) * (sqrt(i) + 1) = sqrt(i) * sqrt(i) + 2 * sqrt(i) + 1

Additional information

Used formulas: (sqrt(i) + 1) * (sqrt(i) + 1) = sqrt(i) * sqrt(i) + 2 * sqrt(i) + 1