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Proof step 28 from proof datsize-write-higher-h (previous step)

n1 + m < # buf, (n1 + m) + 1 < (buf[n1 + m] = empty ⊃ datasize(buf, n1 + m);(n1 + m) + 1), datasize(buf, # buf) < m + n, datasize(buf, # buf) < # buf, m < # buf, n1 < n, n1 < # buf0, datasize(buf, # buf) ≤ n1 + m, datasize(buf, # buf) ≤ m, m + n ≤ # buf, n ≤ # buf0, ∀ 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 ⊦ buf0[n1] = empty

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 29:
n ≠ 0, m ≤ # buf ⊦ ¬ m + n < datasize(buf, m)

Rule argument

pos: left 2

Additional information

Used simplifier rules: datasize-ls-add: n ≠ 0, m ≤ # buf ⊦ ¬ m + n < datasize(buf, m);% ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → m ≤ n; (s)