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Proof step 3 from proof isempty-append (previous step)

⊦ isempty(buf0 + buf1, 0, # buf0 + # buf1) ↔ isempty(buf0, 0, # buf0) ∧ isempty(buf1, 0, # buf1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n0 + n1 ≤ # buf → (isempty(buf, n0, n1) ↔ (∀ n. n0 ≤ n ∧ n < n0 + n1 → buf[n] = empty))
Proof step 5:
m = # buf0 ⊦ isempty(buf0 + buf1, 0, m) ↔ isempty(buf0)
Proof step 6:
⊦ isempty(buf) ↔ isempty(buf, 0, # buf)
Proof step 7:
0 + # buf0 + # buf1 ≤ #(buf0 + buf1) ⊦ (∀ n. 0 ≤ n ∧ n < 0 + # buf0 + # buf1 → (buf0 + buf1)[n] = empty) ↔ isempty(buf0, 0, # buf0) ∧ isempty(buf1, 0, # buf1)

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] → [# buf0 + # buf1, 0, buf0 + buf1] flags: <all paths>

Additional information

Used simplifier rules: isempty-elim-right0: m = # buf0 ⊦ isempty(buf0 + buf1, 0, m) ↔ isempty(buf0);%isempty: ⊦ isempty(buf) ↔ isempty(buf, 0, # buf);% ⊦ m = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ #(buf + buf0) = # buf + # buf0; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s) Used terms: # buf0 + # buf1, 0, buf0 + buf1