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Proof step 5 from proof isempty-elim-right (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n + m ≤ # buf0, n + m ≤ # buf0 + # buf1 ⊦ (∀ n2. n ≤ n2 ∧ n2 < n + m → (buf0 + buf1)[n2] = empty) ↔ isempty(buf0, n, m)

Rule argument

Additional information

Used simplifier rules: append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ #(buf + buf0) = # buf + # buf0; (s)