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Proof step 50 from proof datasize-further (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
buf[n] = empty, n < # buf, n0 < n1, n < n1, n ≤ n0, ∀ n0. n < n0 ∧ n0 < n1 → buf[n0] = empty, ∀ n1, buf. n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)) ⊦ buf[n0] = empty, datasize(buf, n) = datasize(buf, n1)

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → m ≤ n; (s)