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Proof step 8 from proof datasize-at-h (previous step)

buf[m0] = empty, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) < # buf, m0 < # buf, n < # buf, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) ≤ m0, n ≤ m0, datasize(buf, m0) ≤ n, ∀ buf, n. datasize(buf, m0) < # buf ∧ n < m0 ∧ datasize(buf, m0) < m0 ∧ n < # buf ∧ m0 ≤ # buf ∧ datasize(buf, m0) ≤ n → buf[n] = empty ⊦ buf[n] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
buf[m0] = empty, n < # buf, m0 < # buf, datasize(buf, m0) < # buf, datasize(buf, m0) ≤ n, n ≤ m0, datasize(buf, m0) ≤ m0, ∀ buf, n. datasize(buf, m0) < # buf ∧ n < m0 ∧ datasize(buf, m0) < m0 ∧ n < # buf ∧ m0 ≤ # buf ∧ datasize(buf, m0) ≤ n → buf[n] = empty ⊦ buf[n] = empty

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat