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Proof step 5 from proof sqrt-isSquare (previous step)

⊦ abs(sqrt(j * j)) = abs(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ sqrt(i * i) = abs(i)

Rule argument

spec: inst: name: inverse-ii seq: ⊦ sqrt(i * i) = abs(i) subst: [i] → [j] flags: rewrite, current

Additional information

Used simplifier rules: abs-01: ⊦ abs(abs(i)) = abs(i); from specification int-abs Used terms: j