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Proof step 1 from proof square-one-le (conclusion)

⊦ i * i ≤ 1 ↔ i = 0 ∨ i = 1 ∨ i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ i * i ≤ 1 ↔ i = 0 ∨ i = 1 ∨ i + 1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)% ⊦ i = ~1 ↔ i +1 = 0; (s)