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Proof step 26 from proof isSquare-mult (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ isSquare(i) ↔ (∃ j. j * j = i)

Rule argument

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

Additional information

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