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Proof step 16 from proof isempty-elim-left (previous step)

n < (n0 + # buf0) + m, n < # buf0 + # buf1, n0 + # buf0 ≤ n, n0 + m ≤ # buf1, ∀ n. n0 ≤ n ∧ n < n0 + m → buf1[n] = empty ⊦ (n < # buf0 ⊃ buf0[n];buf1[n - # buf0]) = empty, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
n0 < # buf1, n < # buf0 + # buf1, n < (n0 + # buf0) + m, n0 + m ≤ # buf1, n0 + # buf0 ≤ n, ∀ n. n0 ≤ n ∧ n < n0 + m → buf1[n] = empty ⊦ buf1[n - # buf0] = empty, m = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic