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Proof step 4 from proof inverse-ii (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 ≤ i → abs(sqrt(i)) = sqrt(i)
Proof step 6:
sqrt(i * i) < abs(sqrt(i * i) + 1), abs(i) < abs(sqrt(i * i) + 1), sqrt(i * i) ≤ abs(i), 0 ≤ i * i ⊦ sqrt(i * i) = abs(i)

Rule argument

spec: <toplevel> name: root-abs seq: ⊦ 0 ≤ i → abs(sqrt(i)) = sqrt(i) subst: [i] → [i * i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 ≤ i * i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: i * i