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Proof step 54 from proof datasize-further (previous step)

n < # buf, n < n1, n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)), ∀ n0. n < n0 ∧ n0 < n1 → buf[n0] = empty ⊦ n + 1 = datasize(buf, n1), buf[n] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 55:
datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty), n < n1, n < # buf, ∀ n0. n < n0 ∧ n0 < n1 → buf[n0] = empty ⊦ n + 1 = datasize(buf, n1), buf[n] = empty

Rule argument

Additional information

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