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Proof step 18 from proof datsize-write-higher-h (previous step)

buf0[n1] = empty, n1 < # buf0, n1 < n, m < # buf, datasize(buf, # buf) < # buf, datasize(buf, # buf) < m + n, n1 + m < # buf, datasize(copy(buf0, 0, buf, m, n), n1 + m) < (n1 + m) + 1, datasize(buf, # buf) ≤ m, m + n ≤ # buf, n ≤ # buf0, datasize(buf, # buf) ≤ n1 + m, ∀ buf1, buf2, m0, n0. ¬ datasize(copy(buf1, 0, buf2, m0, n0), n1 + m) < datasize(buf2, n1 + m) ∨ ¬ datasize(buf2, # buf2) ≤ m0 ∨ ¬ m0 + n0 ≤ # buf2 ∨ ¬ n0 ≤ # buf1 ∨ ¬ n1 + m ≤ # buf2 ⊦ buf[n1 + m] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
⊦ ¬ n < datasize(buf, n)
Proof step 20:
buf0[n1] = empty, n1 + m < # buf, datasize(buf, # buf) < m + n, datasize(buf, # buf) < # buf, m < # buf, n1 < n, n1 < # buf0, datasize(buf, # buf) ≤ n1 + m, n ≤ # buf0, m + n ≤ # buf, datasize(buf, # buf) ≤ m, ∀ buf1, buf2, m0, n0. ¬ datasize(copy(buf1, 0, buf2, m0, n0), n1 + m) < datasize(buf2, n1 + m) ∨ ¬ datasize(buf2, # buf2) ≤ m0 ∨ ¬ m0 + n0 ≤ # buf2 ∨ ¬ n0 ≤ # buf1 ∨ ¬ n1 + m ≤ # buf2 ⊦ buf[n1 + m] = empty

Rule argument

Additional information

Used simplifier rules: datasize-is-smaller-help: ⊦ ¬ n < datasize(buf, n);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)