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Proof step 6 from proof square-nls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
sqrt(j) < i, i ^2 ≤ j, sqrt(j) ^2 ≤ j, j < (sqrt(j) + 1)^2, 0 ≤ j, i ^2 < (sqrt(j) + 1)^2, sqrt(j) ^2 < (sqrt(j) + 1)^2, 0 < (sqrt(j) + 1)^2 ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%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%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs