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Proof step 17 from proof isempty-elim-left (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
n = n1 + # buf0, n < (n0 + # buf0) + m, n < # buf0 + # buf1, n0 < # buf1, n0 + # buf0 ≤ n, n0 + m ≤ # buf1, # buf0 ≤ n, ∀ n. n0 ≤ n ∧ n < n0 + m → buf1[n] = empty ⊦ buf1[n1] = empty, m = 0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, # buf0, n1]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # buf0, n1