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Proof step 10 from proof isempty-elim-left (previous step)

n0 + m ≤ # buf1 ⊦ (∀ n. n0 + # buf0 ≤ n ∧ n < (n0 + # buf0) + m → (buf0 + buf1)[n] = empty) ↔ (∀ n. n0 ≤ n ∧ n < n0 + m → buf1[n] = empty)

User applied rule case distinction

And obtained the following new goals

Proof step 11:
n0 + m ≤ # buf1, ∀ n. n0 + # buf0 ≤ n ∧ n < (n0 + # buf0) + m → (buf0 + buf1)[n] = empty ⊦ ∀ n. n0 ≤ n ∧ n < n0 + m → buf1[n] = empty
Proof step 13:
n0 + m ≤ # buf1, ∀ n. n0 ≤ n ∧ n < n0 + m → buf1[n] = empty ⊦ ∀ n. n0 + # buf0 ≤ n ∧ n < (n0 + # buf0) + m → (buf0 + buf1)[n] = empty

Rule argument

pos: right 1

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