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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i, sqrt(i) ^2 ≤ i, i < (sqrt(i) + 1)^2, sqrt(i) ^2 < (sqrt(i) + 1)^2, 0 < (sqrt(i) + 1)^2 ⊦ ∃ j0. i = sqrt(i) * sqrt(i) + j0 ∧ j0 ≤ 2 * sqrt(i) ∧ 0 ≤ sqrt(i) ∧ 0 ≤ j0

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%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs