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Proof step 6 from proof square-not-neg-05 (previous step)

sqrt(i) = 0, 1 < (j + 1) * (j + 1), i < (j + 1) * (j + 1), 0 < j, 0 < i, 0 < (j + 1) * (j + 1), j * j ≤ i ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 7:
⊦ 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2
Proof step 8:
sqrt(i) = 0, 0 < (j + 1) * (j + 1), 0 < i, 0 < j, i < (j + 1) * (j + 1), 1 < (j + 1) * (j + 1), j * j ≤ i, 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2 ⊦

Rule argument

spec: <toplevel> name: Sqrt-def seq: ⊦ 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2 subst: [i] → [i]

Additional information

Used terms: i