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Proof step 28 from proof isempty-append (previous step)

n < # buf0 + # buf1, ∀ n. n < # buf1 → buf1[n] = empty, ∀ n. n < # buf0 → buf0[n] = empty ⊦ buf1[n - # buf0] = empty, n < # buf0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
n < # buf0 + # buf1, # buf0 ≤ n, ∀ n. n < # buf1 → buf1[n] = empty, ∀ n. n < # buf0 → buf0[n] = empty ⊦ # buf1 = 0, buf1[n - # buf0] = empty

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)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)