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Proof step 8 from proof isempty-elim-right (previous step)

n + m ≤ # buf0 + # buf1, n + m ≤ # buf0 ⊦ (∀ n2. n ≤ n2 ∧ n2 < n + m → (buf0 + buf1)[n2] = empty) ↔ (∀ n2. n ≤ n2 ∧ n2 < n + m → buf0[n2] = empty)

User applied rule case distinction

And obtained the following new goals

Proof step 9:
n + m ≤ # buf0, n + m ≤ # buf0 + # buf1, ∀ n2. n ≤ n2 ∧ n2 < n + m → (buf0 + buf1)[n2] = empty ⊦ ∀ n2. n ≤ n2 ∧ n2 < n + m → buf0[n2] = empty
Proof step 11:
n + m ≤ # buf0, n + m ≤ # buf0 + # buf1, ∀ n2. n ≤ n2 ∧ n2 < n + m → buf0[n2] = empty ⊦ ∀ n2. n ≤ n2 ∧ n2 < n + m → (buf0 + buf1)[n2] = empty

Rule argument

pos: right 1

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