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Proof step 2 from proof num-i2n-ls-02 (previous step)

0 ≤ i ⊦ m < i→n(i) ↔ n→i(m) < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i; (s)%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs