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Proof step 27 from proof datasize-append-empty-h (previous step)

m0 + 1 = n0 + # buf0, (buf0 + buf1)[m0] = empty, datasize(buf1, n0) = 0, m0 < # buf0 + # buf1, # buf0 ≤ m0, # buf0 ≤ m0 + 1, n0 ≤ # buf1 ⊦ datasize(buf0 + buf1, m0) = datasize(buf0, # buf0), datasize(buf1, m0 - # buf0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
m0 + 1 = n0 + # buf0, buf1[m0 - # buf0] = empty, datasize(buf1, n0) = 0, m0 < # buf0 + # buf1, # buf0 ≤ m0, # buf0 ≤ m0 + 1, n0 ≤ # buf1 ⊦ # buf1 = 0, datasize(buf0 + buf1, m0) = datasize(buf0, # buf0), datasize(buf1, m0 - # buf0) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ # buf ≤ n ∧ n < # buf + # buf0 → (buf + buf0)[n] = buf0[n - # buf]; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic