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Proof step 15 from proof mod-neq (previous step)

i % j = j ⊦ j = 0, 0 < j ∧ 0 ≤ i, j < 0 ∧ 0 ≤ i, 0 < j ∧ i < 0, j < 0 ∧ i < 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ i < j → ¬ j < i; (s)