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Proof step 1 from proof div-one-one-neg (conclusion)
⊦ i | ~1 ↔ i = 1 ∨ i = ~1
User applied rule case distinction
And obtained the following new goals
Proof step 2
:
i | ~1 ⊦ i = 1 ∨ i = ~1
Rule argument
pos: right 1
Additional information
Used simplifier rules: div-one: ⊦ 1 | i;%divides-mone: ⊦ ~1 | j;