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Proof step 1 from proof inverse-ii (conclusion)

⊦ sqrt(i * i) = abs(i)

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2
Proof step 3:
0 ≤ i * i → sqrt(i * i) ^2 ≤ i * i ∧ i * i < sqrt(i * i) +1 ^2 ⊦ sqrt(i * i) = abs(i)

Rule argument

spec: int-square inst: name: Sqrt-def seq: ⊦ 0 ≤ i → sqrt(i) ^2 ≤ i ∧ i < sqrt(i) +1 ^2 subst: [i] → [i * i] flags: current, []

Additional information

Used terms: i * i