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Proof step 11 from proof datasize-ls-samedata-h (previous step)

# buf0 = # buf1, buf0[m0] = empty, m0 < # buf1, datasize(buf0, m0) < m0 + 1, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ ¬ n < datasize(buf, n)
Proof step 13:
# 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

Rule argument

Additional information

Used simplifier rules: datasize-is-smaller-help: ⊦ ¬ n < datasize(buf, n);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)