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Proof step 8 from proof subarray-add-eq-mkempbuf (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n < n0, m + n0 + n1 ≤ # buf, ∀ n. n < n0 + n1 → buf[m + n] = empty ⊦ buf[m + n] = empty

Rule argument

Additional information

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