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Proof step 11 from proof isempty-append (previous step)

⊦ (∀ n. n < # buf0 + # buf1 → (buf0 + buf1)[n] = empty) ↔ (∀ n. n < # buf0 → buf0[n] = empty) ∧ isempty(buf1, 0, # buf1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ n0 + n1 ≤ # buf → (isempty(buf, n0, n1) ↔ (∀ n. n0 ≤ n ∧ n < n0 + n1 → buf[n] = empty))
Proof step 13:
0 + # buf1 ≤ # buf1 ⊦ (∀ n. n < # buf0 + # buf1 → (buf0 + buf1)[n] = empty) ↔ (∀ n. n < # buf0 → buf0[n] = empty) ∧ (∀ n. 0 ≤ n ∧ n < 0 + # buf1 → 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] → [# buf1, 0, buf1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s) Used terms: # buf1, 0, buf1