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Proof step 5 from proof num-i2n-ls (previous step)

i < n→i(m), 0 ≤ i, m ≤ i→n(i) ⊦ m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n

Rule argument

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

Additional information

Used simplifier rules: i2i-pos-01: ⊦ 0 ≤ i → n→i(i→n(i)) = i;% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: i→n(i), m