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Proof step 27 from proof datsize-write-higher-h (previous step)

m1 = n1 + m, datasize(buf, # buf) < # buf, datasize(buf, # buf) < m + n, m1 + 1 < (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1), m1 < # buf, m1 < m + n, m < # buf, n ≤ # buf0, m + n ≤ # buf, datasize(buf, # buf) ≤ m, m ≤ m1, datasize(buf, # buf) ≤ m1, ∀ 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 ⊦ buf0[n1] = empty, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
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

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (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)