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Proof step 2 from proof isempty-elim-left0 (previous step)

m ≤ # buf1 ⊦ isempty(buf0 + buf1, # buf0, m) ↔ isempty(buf1, 0, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ n0 + n1 ≤ # buf → (isempty(buf, n0, n1) ↔ (∀ n. n0 ≤ n ∧ n < n0 + n1 → buf[n] = empty))
Proof step 4:
m ≤ # buf1, # buf0 + m ≤ #(buf0 + buf1) ⊦ (∀ n. # buf0 ≤ n ∧ n < # buf0 + m → (buf0 + buf1)[n] = empty) ↔ isempty(buf1, 0, m)

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

Additional information

Used simplifier rules: ⊦ #(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)% ⊦ m ≤ n → (n < m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m, # buf0, buf0 + buf1