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Proof step 10 from proof isSquare-mult (previous step)

abs(j0) * abs(j0) = abs(j) * abs(i) * abs(i), j0 * j0 = j * i * i, isSquare(abs(j)) ⊦ i = 0, isSquare(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ isSquare(i) ↔ (∃ j. j * j = i)
Proof step 12:
abs(j0) * abs(j0) = abs(j) * abs(i) * abs(i), j0 * j0 = j * i * i, ∃ j0. j0 * j0 = abs(j) ⊦ i = 0, isSquare(j)

Rule argument

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

Additional information

Used terms: abs(j)