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Proof step 3 from proof mone-07 (previous step)

⊦ ¬ i < ~1 ↔ ¬ i < 0, ~1 = i

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: mone-06: ⊦ ~1 < i ↔ 0 ≤ i;%lels-03: ⊦ i < j → (j ≤ i ↔ false);%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)