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Proof step 3 from proof isSquare-mult (previous step)

isSquare(j * i * i) ⊦ i = 0, isSquare(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ isSquare(i) ↔ (∃ j. j * j = i)
Proof step 5:
∃ j0. j0 * j0 = j * i * i ⊦ i = 0, isSquare(j)

Rule argument

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

Additional information

Used terms: j * i * i