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Proof step 12 from proof two-le-three (previous step)

1 < i ^2, 0 < i ^2, 3 < i ^2, abs(i) ≤ 1, abs(i) < i ^2 ⊦ i ≠ 0 ∧ i ≠ 1

User applied rule case distinction

And obtained the following new goals

Proof step 13:
1 < i ^2, 0 < i ^2, 3 < i ^2, abs(i) ≤ 1, abs(i) < i ^2 ⊦ i ≠ 0
Proof step 16:
1 < i ^2, 0 < i ^2, 3 < i ^2, abs(i) ≤ 1, abs(i) < i ^2 ⊦ i ≠ 1

Rule argument

pos: right 1

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