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Proof step 4 from proof isempty-elim-left0 (previous step)

m ≤ # buf1, # buf0 + m ≤ #(buf0 + buf1) ⊦ (∀ n. # buf0 ≤ n ∧ n < # buf0 + m → (buf0 + buf1)[n] = empty) ↔ isempty(buf1, 0, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m ≤ # buf1 ⊦ (∀ n. # buf0 ≤ n ∧ n < # buf0 + m → (buf0 + buf1)[n] = empty) ↔ isempty(buf1, 0, m)

Rule argument

Additional information

Used simplifier rules: 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)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ #(buf + buf0) = # buf + # buf0; (s)