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Proof step 17 from proof datasize-ls-samedata-h (previous step)

# 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ range=(buf, n0, buf0, n1, m) ↔ n0 + m ≤ # buf ∧ n1 + m ≤ # buf0 ∧ (∀ n. n < m → buf[n0 + n] = buf0[n1 + n])
Proof step 19:
# buf0 = # buf1, buf0[m0] = empty, m0 < # buf1, m0 < datasize(buf1, # buf1), 0 + datasize(buf1, # buf1) ≤ # buf0, 0 + datasize(buf1, # buf1) ≤ # buf1, ∀ buf, buf2. # buf ≠ # buf2 ∨ ¬ datasize(buf, m0) < datasize(buf2, m0) ∨ ¬ range=(buf, 0, buf2, 0, datasize(buf2, # buf2)) ∨ ¬ m0 ≤ # buf2, ∀ n. n < datasize(buf1, # buf1) → buf0[0 + n] = buf1[0 + n] ⊦ buf1[m0] = empty

Rule argument

spec: array-splice, inst: buffer name: range-eq-ext seq: ⊦ range=(buf, n0, buf0, n1, m) ↔ n0 + m ≤ # buf ∧ n1 + m ≤ # buf0 ∧ (∀ n. n < m → buf[n0 + n] = buf0[n1 + n]) subst: [m, n1, buf0, n0, buf] → [datasize(buf1, # buf1), 0, buf1, 0, buf0] flags: <all paths>

Additional information

Used terms: datasize(buf1, # buf1), 0, buf1, 0, buf0