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Proof step 8 from proof isempty-elim-left (previous step)

n0 + m ≤ # buf1 ⊦ (∀ n. n0 + # buf0 ≤ n ∧ n < (n0 + # buf0) + m → (buf0 + buf1)[n] = empty) ↔ isempty(buf1, n0, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ n0 + n1 ≤ # buf → (isempty(buf, n0, n1) ↔ (∀ n. n0 ≤ n ∧ n < n0 + n1 → buf[n] = empty))
Proof step 10:
n0 + m ≤ # buf1 ⊦ (∀ n. n0 + # buf0 ≤ n ∧ n < (n0 + # buf0) + m → (buf0 + buf1)[n] = empty) ↔ (∀ n. n0 ≤ n ∧ n < n0 + m → buf1[n] = 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, n0, buf1] flags: <all paths>

Additional information

Used terms: m, n0, buf1