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Proof step 22 from proof datasize-add-empty-h (previous step)

# subarray(buf, n0, n + 1) = # mkempbuf(n + 1), n0 + n < # buf, ∀ buf, n0. subarray(buf, n0, n) = mkempbuf(n) ∧ n0 + n ≤ # buf → datasize(buf, n0 + n) = datasize(buf, n0), ∀ n1. n1 < # subarray(buf, n0, n + 1) → subarray(buf, n0, n + 1)[n1] = mkempbuf(n + 1)[n1] ⊦ (n0 + n) + 1 = datasize(buf, n0), buf[n0 + n] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
n0 + n < # buf, ∀ buf, n0. subarray(buf, n0, n) = mkempbuf(n) ∧ n0 + n ≤ # buf → datasize(buf, n0 + n) = datasize(buf, n0), ∀ n1. ¬ n < n1 → subarray(buf, n0, n + 1)[n1] = mkempbuf(n + 1)[n1] ⊦ (n0 + n) + 1 = datasize(buf, n0), buf[n0 + n] = empty

Rule argument

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)%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)% ⊦ # mkempbuf(n) = n; (s)% ⊦ # subarray(buf, m, n) = n; (s)