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Proof step 10 from proof rangeeq-emptyleb-datasize (previous step)

# buf = n0 + n1, n2 < n0 + n1, range=(buf, n0, emptyleb, 0, n1), n0 ≤ n2, ∀ buf0, n, n3. # buf0 ≠ n + n3 ∨ ¬ n < datasize(buf0, n2) ∨ ¬ range=(buf0, n, emptyleb, 0, n3) ∨ ¬ n2 ≤ n + n3 ⊦ buf[n2] = empty, n1 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ range=(buf, n0, buf0, n1, m) ↔ n0 + m ≤ # buf ∧ n1 + m ≤ # buf0 ∧ (∀ n. n < m → buf[n0 + n] = buf0[n1 + n])
Proof step 12:
# buf = n0 + n1, n2 < n0 + n1, n0 ≤ n2, n0 + n1 ≤ # buf, 0 + n1 ≤ # emptyleb, ∀ buf0, n, n3. # buf0 ≠ n + n3 ∨ ¬ n < datasize(buf0, n2) ∨ ¬ range=(buf0, n, emptyleb, 0, n3) ∨ ¬ n2 ≤ n + n3, ∀ n. n < n1 → buf[n0 + n] = (emptyleb)[0 + n] ⊦ buf[n2] = empty, n1 = 0

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] → [n1, 0, emptyleb, n0, buf] flags: <all paths>

Additional information

Used terms: n1, 0, emptyleb, n0, buf