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Proof step 1 from proof isSquare-square (conclusion)

⊦ isSquare(dec * dec)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ isSquare(dec_i) ↔ (∃ dec. dec * dec = dec_i)
Proof step 3:
⊦ ∃ dec0. dec0 * dec0 = dec * dec

Rule argument

spec: inst: name: isSquare-def seq: ⊦ isSquare(dec_i) ↔ (∃ dec. dec * dec = dec_i) subst: [dec_i] → [dec * dec] flags: rewrite, current

Additional information

Used terms: dec * dec