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Proof step 1 from proof le-int-le-nat (conclusion)
⊦ i ≥ 0 → ((* i) + n→i(n) ≤ n→i(m) ↔ i→n(i) + n ≤ m)
Special step normalize
And obtained the following new goals
Proof step 2
:
⊦ i ≥ 0 → (i + n→i(n) ≤ n→i(m) ↔ i→n(i) + n ≤ m)