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Proof step 64 from proof datasize-further (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
buf[n0] = empty, n < # buf, n ≤ n2, n2 ≤ n0, n ≤ n0, ∀ n1. n < n1 ∧ ¬ n0 < n1 → buf[n1] = empty ⊦ n + 1 = datasize(buf, n0), buf[n2] = empty, buf[n] = empty, datasize(buf, n) = datasize(buf, n0), ¬(∀ n2. n ≤ n2 ∧ n2 < n0 → buf[n2] = empty) ∧ n < n0 ∧ (n < n1 ∧ n1 < n0 → buf[n1] = empty)

Rule argument

Additional information

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