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Proof step 12 from proof datsize-write-higher-h (previous step)

buf0[m1 - m] = empty, datasize(copy(buf0, 0, buf, m, n), m1) < (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 ⊦ m1 < m, m + n ≤ m1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
buf0[m1 - m] = empty, m < # buf, m1 < m + n, datasize(copy(buf0, 0, buf, m, n), m1) < (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1), m1 < # buf, datasize(buf, # buf) < m + n, datasize(buf, # buf) < # buf, datasize(buf, # buf) ≤ m1, m ≤ m1, 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 ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n ↔ ¬ n < m; (ss)