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Proof step 11 from proof isempty-elim-right (previous step)

n + m ≤ # buf0, n + m ≤ # buf0 + # buf1, ∀ n2. n ≤ n2 ∧ n2 < n + m → buf0[n2] = empty ⊦ ∀ n2. n ≤ n2 ∧ n2 < n + m → (buf0 + buf1)[n2] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n < # buf0 + # buf1, n < # buf0, n0 < # buf0 + # buf1, n0 < # buf0, n0 < n + m, n ≤ n0, n + m ≤ # buf0 + # buf1, n + m ≤ # buf0, ∀ n2. n ≤ n2 ∧ n2 < n + m → buf0[n2] = empty ⊦ buf0[n0] = empty, m = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ n < # buf → (buf + buf0)[n] = buf[n]; (s)% ⊦ m < n → n ≠ 0; (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