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Proof step 2 from proof num-i2n-le (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 3:
0 ≤ i, i ≤ n→i(m) ⊦ i→n(i) ≤ m

Rule argument

pos: right 1

Additional information

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