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Proof step 14 from proof labs (previous step)

⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j), 0 ≤ i ∧ 0 ≤ j, 0 ≤ i ∧ j ≤ 0, i ≤ 0 ∧ 0 ≤ j, i ≤ 0 ∧ j ≤ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 5

Additional information

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