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

m0 ≤ # buf, m0 ≤ # buf ∧ m ≠ 0 → (m ≤ datasize(buf, m0) ↔ (∃ n. n + 1 ≥ m ∧ n < m0 ∧ buf[n] ≠ empty)) ⊦ datasize(buf, m0) < m ↔ ¬(∃ n. m ≤ n + 1 ∧ n < m0 ∧ buf[n] ≠ empty), m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ datasize(buf, m) ≤ n ∧ n < m ∧ m ≤ # buf → buf[n] = empty
Proof step 7:
n < m0, n < # buf, m ≤ # buf, m0 ≤ # buf, m ≤ datasize(buf, m0), m ≤ n + 1, m ≤ m0 ⊦ buf[n] = empty, m = 0, ∃ n. m ≤ n + 1 ∧ n < m0 ∧ buf[n] ≠ empty

Rule argument

Additional information

Used simplifier rules: datasize-at-h: ⊦ datasize(buf, m) ≤ n ∧ n < m ∧ m ≤ # buf → buf[n] = empty;% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ 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%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ n ≥ m ↔ m ≤ n; (s)