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Proof step 5 from proof isempty-ext (previous step)

n0 + n +1 ≤ # buf, ∀ n0, buf. n0 + n ≤ # buf → (isempty(buf, n0, n) ↔ (∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty)) ⊦ isempty(buf, n0, n +1) ↔ (∀ n2. n0 ≤ n2 ∧ n2 < n0 + n +1 → buf[n2] = empty)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 + n < # buf, ∀ n0, buf. n0 + n ≤ # buf → (isempty(buf, n0, n) ↔ (∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty)) ⊦ isempty(buf, n0, n + 1) ↔ (∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty)

Rule argument

Additional information

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