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Proof step 7 from proof isempty-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ (∀ n. n < # buf0 + # buf1 → (buf0 + buf1)[n] = empty) ↔ isempty(buf0, 0, # buf0) ∧ isempty(buf1, 0, # buf1)

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s)% ⊦ 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)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ 0 ≤ m; (s)