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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2