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Proof step 3 from proof num-i2n-ls (previous step)

0 ≤ i ⊦ i→n(i) < m ↔ i < n→i(m)

User applied rule case distinction

And obtained the following new goals

Proof step 4:
i < n→i(m), 0 ≤ i ⊦ i→n(i) < m

Rule argument

pos: right 1

Additional information

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