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Proof step 5 from proof sqrt-mult-squares (previous step)

isSquare(i) ⊦ sqrt(i * j0 * j0) = sqrt(i) * sqrt(j0 * j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ isSquare(i) ↔ (∃ j. j * j = i)
Proof step 7:
∃ j. j * j = i ⊦ sqrt(i * j0 * j0) = sqrt(i) * sqrt(j0 * j0)

Rule argument

spec: inst: name: isSquare-def seq: ⊦ isSquare(i) ↔ (∃ j. j * j = i) subst: [i] → [i] flags: rewrite, current

Additional information

Used terms: i