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Proof step 4 from proof sqrt-estimate (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: two-not-neg: ⊦ ¬ i ^2 < 0;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2