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Proof step 6 from proof isempty-elim-right (previous step)

n + m ≤ # buf0, n + m ≤ # buf0 + # buf1 ⊦ (∀ n2. n ≤ n2 ∧ n2 < n + m → (buf0 + buf1)[n2] = empty) ↔ isempty(buf0, n, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n0 + n1 ≤ # buf → (isempty(buf, n0, n1) ↔ (∀ n. n0 ≤ n ∧ n < n0 + n1 → buf[n] = empty))
Proof step 8:
n + m ≤ # buf0 + # buf1, n + m ≤ # buf0 ⊦ (∀ n2. n ≤ n2 ∧ n2 < n + m → (buf0 + buf1)[n2] = empty) ↔ (∀ n2. n ≤ n2 ∧ n2 < n + m → buf0[n2] = empty)

Rule argument

spec: <toplevel> name: isempty-ext seq: ⊦ n0 + n1 ≤ # buf → ( isempty(buf, n0, n1) ↔ (∀ n. n0 ≤ n ∧ n < n0 + n1 → buf[n] = empty)) subst: [n1, n0, buf] → [m, n, buf0] flags: <all paths>

Additional information

Used terms: m, n, buf0