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Proof step 18 from proof datasize-empbuf-h (previous step)

mkempbuf(n)[n0] = empty, m0 = n0 + # buf, m0 < # buf + n, # buf ≤ m0, ∀ buf, n. m0 ≤ # buf + n → datasize(buf + mkempbuf(n), m0) = datasize(buf, min(# buf, m0)) ⊦ datasize(buf + mkempbuf(n), m0) = datasize(buf, # buf), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
n0 < n, ∀ buf0, n. n0 + # buf ≤ # buf0 + n → datasize(buf0 + mkempbuf(n), n0 + # buf) = datasize(buf0, min(# buf0, n0 + # buf)) ⊦ datasize(buf + mkempbuf(n), n0 + # buf) = datasize(buf, # buf)

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → mkempbuf(n)[m] = empty; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)