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Proof step 12 from proof rangeeq-emptyleb-datasize (previous step)

# buf = n0 + n1, n2 < n0 + n1, n0 ≤ n2, n0 + n1 ≤ # buf, 0 + n1 ≤ # emptyleb, ∀ buf0, n, n3. # buf0 ≠ n + n3 ∨ ¬ n < datasize(buf0, n2) ∨ ¬ range=(buf0, n, emptyleb, 0, n3) ∨ ¬ n2 ≤ n + n3, ∀ n. n < n1 → buf[n0 + n] = (emptyleb)[0 + n] ⊦ buf[n2] = empty, n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# buf = n0 + n1, n2 < n0 + n1, n1 ≤ LEB_SIZE, n0 ≤ n2, ∀ n. n < n1 → buf[n0 + n] = empty, ∀ buf0, n, n3. # buf0 ≠ n + n3 ∨ ¬ n < datasize(buf0, n2) ∨ ¬ range=(buf0, n, emptyleb, 0, n3) ∨ ¬ n2 ≤ n + n3 ⊦ buf[n2] = empty, n1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%n < # emptyleb ⊦ (emptyleb)[n] = empty; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # emptyleb = LEB_SIZE; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)