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Proof step 4 from proof num-n2i-ls (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

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

Rule argument

spec: <toplevel> name: num-i2n-le seq: 0 ≤ i ⊦ i→n(i) ≤ (* m) ↔ i ≤ n→i(m) subst: [m, i] → [n, i] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: n, i