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Proof step 3 from proof isempty-elim-left (previous step)

n = n0 + # buf0, # buf0 ≤ n, n + m ≤ # buf0 + # buf1 ⊦ isempty(buf0 + buf1, n, m) ↔ isempty(buf1, n0, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 + m ≤ # buf1 ⊦ isempty(buf0 + buf1, n0 + # buf0, m) ↔ isempty(buf1, n0, m)

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (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)