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Proof step 3 from proof square-not-neg-05 (previous step)

sqrt(i) = 0, 0 < i ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → (∃ j. 0 < j ∧ j * j ≤ i ∧ i < (j + 1) * (j + 1))
Proof step 5:
sqrt(i) = 0, 0 < i, 0 < i → (∃ j. 0 < j ∧ j * j ≤ i ∧ i < (j + 1) * (j + 1)) ⊦

Rule argument

spec: int-mult name: square-ex seq: ⊦ 0 < i → (∃ j. 0 < j ∧ j * j ≤ i ∧ i < (j + 1) * (j + 1)) subst: [i] → [i]

Additional information

Used terms: i