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Proof step 9 from proof ls-datasize-zero (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# buf = n1 + n, n0 < n1 + n, ∀ n0. n0 < n → buf[n0] = empty ⊦ buf[n0] = (mkempbuf(n) + subarray(buf, n, n1))[n0]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ # subarray(buf, m, n) = n; (s)% ⊦ # mkempbuf(n) = n; (s)% ⊦ #(buf + buf0) = # buf + # buf0; (s)