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Proof step 2 from proof square-one-le (previous step)

⊦ i * i ≤ 1 ↔ i = 0 ∨ i = 1 ∨ i + 1 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
i * i ≤ 1 ⊦ i + 1 = 0, i = 0, i = 1
Proof step 7:
i = 0 ∨ i = 1 ∨ i + 1 = 0 ⊦ i * i ≤ 1

Rule argument

pos: right 1

Additional information