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Proof step 3 from proof num-i2n-ls-02 (previous step)

0 ≤ i ⊦ m < i→n(i) ↔ n→i(m) < i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n→i(m) < n→i(n) ↔ m < n
Proof step 5:
0 ≤ i ⊦ n→i(m) < n→i(i→n(i)) ↔ n→i(m) < i

Rule argument

spec: <toplevel> name: nat2int-04 seq: ⊦ n→i(m) < n→i(n) ↔ m < n subst: [n, m] → [i→n(i), m] flags: <rotated> <all paths>

Additional information

Used terms: i→n(i), m