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Proof step 18 from proof pot-isSquare (previous step)

isSquare(i), 0 < i, 0 < j ⊦ isSquare(i ^ j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ isSquare(i) ↔ (∃ j. j * j = i)
Proof step 20:
∃ j. j * j = i, 0 < i, 0 < j ⊦ isSquare(i ^ j)

Rule argument

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

Additional information

Used terms: i