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Proof step 12 from proof isempty-ext (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
n0 + n < # buf, n1 < n0 + n, n1 + 1 < # buf, n1 < # buf, n0 < # buf, n0 ≤ n1, ∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty ⊦ buf[n1] = empty, n = 0, isempty(buf, n0, n)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic