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Proof step 2 from proof num-n2i-ls (previous step)

0 ≤ i ⊦ n→i(n) < i ↔ n < i→n(i)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
n < i→n(i), 0 ≤ i ⊦ n→i(n) < i

Rule argument

pos: right 1

Additional information

Used simplifier rules: inni-le: ⊦ i→n(i) ≤ n → ¬ n→i(n) < i;% ⊦ m < n ↔ ¬ n ≤ m; (ss)