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Proof step 15 from proof datasize-ls-samedata-h (previous step)

# buf0 = # buf1, buf0[m0] = empty, m0 < # buf1, range=(buf0, 0, buf1, 0, datasize(buf1, # buf1)), ∀ buf, buf2. # buf ≠ # buf2 ∨ ¬ datasize(buf, m0) < datasize(buf2, m0) ∨ ¬ range=(buf, 0, buf2, 0, datasize(buf2, # buf2)) ∨ ¬ m0 ≤ # buf2 ⊦ buf1[m0] = empty, datasize(buf1) ≤ m0 ∧ m0 < # buf1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
⊦ datasize(buf) = datasize(buf, # buf)
Proof step 17:
# buf0 = # buf1, buf0[m0] = empty, m0 < datasize(buf1, # buf1), m0 < # buf1, range=(buf0, 0, buf1, 0, datasize(buf1, # buf1)), ∀ buf, buf2. # buf ≠ # buf2 ∨ ¬ datasize(buf, m0) < datasize(buf2, m0) ∨ ¬ range=(buf, 0, buf2, 0, datasize(buf2, # buf2)) ∨ ¬ m0 ≤ # buf2 ⊦ buf1[m0] = empty

Rule argument

Additional information

Used simplifier rules: datasize: ⊦ datasize(buf) = datasize(buf, # buf);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)