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Proof step 15 from proof isempty-elim-left0 (previous step)

n < # buf0 + m, n < # buf0 + # buf1, # buf0 ≤ n, m ≤ # buf1, ∀ n. n < m → buf1[n] = empty ⊦ # buf1 = 0, buf1[n - # buf0] = empty, m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
n = n0 + # buf0, n < # buf0 + # buf1, n < # buf0 + m, m ≤ # buf1, # buf0 ≤ n, ∀ n. n < m → buf1[n] = empty ⊦ # buf1 = 0, buf1[n0] = 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, n0]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # buf0, n0