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Proof step 3 from proof square-nls (previous step)

sqrt(j) < i, i ^2 ≤ j ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2
Proof step 5:
sqrt(j) < i, i ^2 ≤ j, 0 ≤ j → sqrt(j) ^2 ≤ j ∧ j < sqrt(j) +1 ^2 ⊦

Rule argument

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

Additional information

Used terms: j