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Proof step 6 from proof square-one-le (previous step)

i * i ≤ 1 * 1 ↔ abs(i) ≤ abs(1), i * i ≤ 1 ⊦ i + 1 = 0, i = 0, i = 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)% ⊦ i = ~1 ↔ i +1 = 0; (s)% ⊦ i ≠ 0 ∧ i ≠ 1 → (abs(i) ≤ 1 ↔ i = ~1); (s)