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Proof step 16 from proof isempty-ext (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
buf[n0 + n] = empty, n1 < # buf, n0 + n < # buf, n0 < # buf, isempty(buf, n0, n), n1 ≤ n0 + n, n0 ≤ n1, ∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty ⊦ buf[n1] = empty

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; (s)%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)