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Proof step 3 from proof ls-int-ls-nat (previous step)

0 ≤ i ⊦ i + n→i(n) < n→i(m) ↔ i→n(i) + n < m

User applied rule case distinction

And obtained the following new goals

Proof step 4:
i + n→i(n) < n→i(m), 0 ≤ i ⊦ i→n(i) + n < m
Proof step 7:
i→n(i) + n < m, 0 ≤ i ⊦ i + n→i(n) < n→i(m)

Rule argument

pos: right 1

Additional information