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Proof step 3 from proof subarray-append-mkempbuf-from (previous step)

n2 ≤ n1 + # buf ⊦ # subarray(mkempbuf(n0 + n1) + buf, n0, n2) = # subarray(mkempbuf(n1) + buf, 0, n2) ∧ (∀ n. n < # subarray(mkempbuf(n0 + n1) + buf, n0, n2) → subarray(mkempbuf(n0 + n1) + buf, n0, n2)[n] = subarray(mkempbuf(n1) + buf, 0, n2)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n < n2, n < n1 + # buf, n2 ≤ n1 + # buf ⊦ (mkempbuf(n0 + n1) + buf)[n0 + n] = (mkempbuf(n1) + buf)[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ n0 < n → subarray(buf, m, n)[n0] = buf[m + n0]; (s)% ⊦ # subarray(buf, m, n) = n; (s)