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

⊦ i * i ≤ 1 ↔ i = 0 ∨ i = 1 ∨ i = ~1

User applied rule case distinction

And obtained the following new goals

Proof step 3:
i * i ≤ 1 ⊦ i = 0 ∨ i = 1 ∨ i = ~1

Rule argument

pos: right 1

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs