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Proof step 4 from proof num-i2n-le (previous step)

i ≤ n→i(m), m < i→n(i), 0 < i ⊦ i = 0, m = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ n→i(m) < n→i(n) ↔ m < n

Rule argument

spec: inst: name: nat2int-04 seq: ⊦ n→i(m) < n→i(n) ↔ m < n subst: [m, n] → [m, i→n(i)] flags: current, []

Additional information

Used simplifier rules: i2i-pos: ⊦ 0 < i → n→i(i→n(i)) = i;%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs Used terms: m, i→n(i)