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Proof step 1 from proof isempty-elim-right0 (conclusion)

m = # buf0 ⊦ isempty(buf0 + buf1, 0, m) ↔ isempty(buf0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n + m ≤ # buf0, n + m ≤ # buf0 + # buf1 ⊦ isempty(buf0 + buf1, n, m) ↔ isempty(buf0, n, m)
Proof step 3:
⊦ isempty(buf) ↔ isempty(buf, 0, # buf)

Rule argument

Additional information

Used simplifier rules: isempty-elim-right: n + m ≤ # buf0, n + m ≤ # buf0 + # buf1 ⊦ isempty(buf0 + buf1, n, m) ↔ isempty(buf0, n, m);%isempty: ⊦ isempty(buf) ↔ isempty(buf, 0, # buf);%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ m ≤ m + n; (s)% ⊦ m ≤ m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)