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Proof step 1 from proof square-ex (conclusion)

⊦ 0 < i → (∃ j. 0 < j ∧ j * j ≤ i ∧ i < (j + 1) * (j + 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < i ⊦ ∃ j. 0 < j ∧ j * j ≤ i ∧ i < (j + 1) * (j + 1)

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs