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Proof step 12 from proof root-elim (previous step)

i < sqrt(i) * sqrt(i) + 2 * sqrt(i) + 1, sqrt(i) * sqrt(i) ≤ i, 0 ≤ i, 0 < sqrt(i) * sqrt(i) + 2 * 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: root-abs: ⊦ 0 ≤ i → abs(sqrt(i)) = sqrt(i);%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%abs-04: ⊦ 0 ≤ i → abs(i + 1) = abs(i) + 1; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; 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%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs