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Proof step 2 from proof isempty-elim-left (previous step)

n + m ≤ # buf0 + # buf1, # buf0 ≤ n ⊦ isempty(buf0 + buf1, n, m) ↔ isempty(buf1, n - # buf0, m)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = n0 + # buf0, # buf0 ≤ n, n + m ≤ # buf0 + # buf1 ⊦ isempty(buf0 + buf1, n, m) ↔ isempty(buf1, n0, m)

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