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

m < i→n(i), 0 < i, i ≤ n→i(m) ⊦ 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: <toplevel> name: nat2int-04 seq: ⊦ n→i(m) < n→i(n) ↔ m < n subst: [m, n] → [m, i→n(i)]

Additional information

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