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Proof step 6 from proof isempty-ext (previous step)

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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n + m < # buf → (isempty(buf, n, m + 1) ↔ buf[n + m] = empty ∧ isempty(buf, n, m))
Proof step 8:
n0 + n < # buf, ∀ n0, buf. n0 + n ≤ # buf → (isempty(buf, n0, n) ↔ (∀ n2. n0 ≤ n2 ∧ n2 < n0 + n → buf[n2] = empty)) ⊦ buf[n0 + n] = empty ∧ isempty(buf, n0, n) ↔ (∀ n1. n0 ≤ n1 ∧ ¬ n0 + n < n1 → buf[n1] = empty)

Rule argument

spec: <toplevel> name: is-empty-range-rec seq: ⊦ n + m < # buf → ( isempty(buf, n, m + 1) ↔ buf[n + m] = empty ∧ isempty(buf, n, m)) subst: [m, n, buf] → [n, n0, buf] flags: <all paths>

Additional information

Used terms: n, n0, buf