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Proof step 4 from proof root-nls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
sqrt(i) ^2 < sqrt(i) * sqrt(i), i < sqrt(i) ^2 + 2 * sqrt(i) + 1 ^2, sqrt(i) ^2 ≤ i, 0 < sqrt(i) * sqrt(i), 0 ≤ i, i < sqrt(i) * sqrt(i), 0 < sqrt(i) ^2 + 2 * sqrt(i) + 1 ^2, 0 < 2 * sqrt(i) + 1 ^2 ⊦

Rule argument

Additional information

Used simplifier rules: two-plus: ⊦ (i + j)^2 = i ^2 + 2 * i * j + j ^2;%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%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%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs