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Proof step 43 from proof datasize-further (previous step)

n + 1 = datasize(buf, n2), buf[n2] = empty, n < # buf, n < n0, n < n2, n0 ≤ n2, datasize(buf, n2) ≤ # buf, datasize(buf, n2) ≤ n0, datasize(buf, n2) ≤ n2 ⊦ datasize(buf, n) = datasize(buf, n2) ↔ (∀ n3. n ≤ n3 ∧ n3 < n2 → buf[n3] = empty), buf[n] = empty, buf[n0] = empty, datasize(buf, n) = datasize(buf, n2), ∀ n1. n ≤ n1 ∧ ¬ n2 < n1 → buf[n1] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
n + 1 = datasize(buf, n2), buf[n2] = empty, n < # buf, n < n0, n < n2, n1 ≤ n2, n0 ≤ n2, datasize(buf, n2) ≤ # buf, datasize(buf, n2) ≤ n0, datasize(buf, n2) ≤ n2, n ≤ n1, ∀ n3. n ≤ n3 ∧ n3 < n2 → buf[n3] = empty ⊦ buf[n] = empty, buf[n0] = empty, buf[n1] = empty, datasize(buf, n) = datasize(buf, n2)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)