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Proof step 5 from proof root-nls (previous step)

0 < sqrt(i) * sqrt(i), i < sqrt(i) * sqrt(i), 0 ≤ i, 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i < sqrt(i) * sqrt(i), 0 < sqrt(i) * sqrt(i), i < (sqrt(i) + 1)^2, sqrt(i) ^2 < sqrt(i) * sqrt(i), sqrt(i) ^2 < (sqrt(i) + 1)^2, 0 < (sqrt(i) + 1)^2, 0 ≤ i, sqrt(i) ^2 ≤ i ⊦

Rule argument

Additional information

Used simplifier rules: forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i +1 = i + 1; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)