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Proof step 10 from proof isempty-ext (previous step)

n0 + n < # buf, n0 + n ≤ # buf → (isempty(buf, n0, n) ↔ (∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty)) ⊦ buf[n0 + n] = empty ∧ isempty(buf, n0, n) ↔ (∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
isempty(buf, n0, n) ↔ (∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty), n0 + n < # buf ⊦ buf[n0 + n] = empty ∧ isempty(buf, n0, n) ↔ (∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)