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Proof step 5 from proof datasize-at-h (previous step)

n < # buf, datasize(buf, m0 +1) < m0 +1, n < m0 +1, datasize(buf, m0 +1) < # buf, datasize(buf, m0 +1) ≤ n, m0 +1 ≤ # buf, ∀ 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 6:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 7:
n < # buf, m0 < # buf, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) < # buf, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) ≤ n, n ≤ m0, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) ≤ 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: datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)