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Proof step 6 from proof num-i2n-ls-02 (previous step)

0 ≤ i ⊦ n→i(m) < abs(i) ↔ n→i(m) < i

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

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