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Proof step 1 from proof isSquare-square (conclusion)
⊦ isSquare(i * i)
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 2
:
⊦ isSquare(i) ↔ (∃ j. j * j = i)
Proof step 3
:
⊦ ∃ j. j * j = i * i
Rule argument
spec: inst: name: isSquare-def seq: ⊦ isSquare(i) ↔ (∃ j. j * j = i) subst: [i] → [i * i] flags: rewrite, current
Additional information
Used terms: i * i