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Proof step 6 from proof datsize-write-higher-h (previous step)

datasize(copy(buf0, 0, buf, m, n), m1 +1) < datasize(buf, m1 +1), m1 +1 ≤ # buf, n ≤ # buf0, m + n ≤ # buf, datasize(buf, # buf) ≤ m, ∀ buf0, buf, m, n. ¬ (datasize(copy(buf0, 0, buf, m, n), m1) < datasize(buf, m1) ∧ datasize(buf, # buf) ≤ m ∧ m + n ≤ # buf ∧ n ≤ # buf0 ∧ m1 ≤ # buf) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 8:
((m1 < m ∨ m + n ≤ m1 ⊃ buf[m1];buf0[m1 - m]) = empty ⊃ datasize(copy(buf0, 0, buf, m, n), m1);m1 + 1) < (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1), m1 < # buf, n ≤ # buf0, m + n ≤ # buf, datasize(buf, # buf) ≤ m, ∀ buf1, buf2, m0, n0. ¬ datasize(copy(buf1, 0, buf2, m0, n0), m1) < datasize(buf2, m1) ∨ ¬ datasize(buf2, # buf2) ≤ m0 ∨ ¬ m0 + n0 ≤ # buf2 ∨ ¬ n0 ≤ # buf1 ∨ ¬ m1 ≤ # buf2 ⊦

Rule argument

Additional information

Used simplifier rules: datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n + 0 = n; (s)%⊦ m0 < # buf → copy(buf0, n, buf, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ buf[m0];buf0[m0 + n - n0]); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)