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Proof step 5 from proof rangeeq-subarray-5 (previous step)

n0 < m0, range=(avol, 5 + n, avol0, 5 + n, m0) ⊦ avol[(5 + n) + n0] = avol0[(5 + n) + n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ range=(avol, n0, avol0, n1, m) ↔ n0 + m ≤ # avol ∧ n1 + m ≤ # avol0 ∧ (∀ n. n < m → avol[n0 + n] = avol0[n1 + n])
Proof step 7:
n0 < m0, (5 + n) + m0 ≤ # avol, (5 + n) + m0 ≤ # avol0, ∀ n2. n2 < m0 → avol[(5 + n) + n2] = avol0[(5 + n) + n2] ⊦ avol[(5 + n) + n0] = avol0[(5 + n) + n0]

Rule argument

spec: array-splice, inst: avolume name: range-eq-ext seq: ⊦ range=(avol, n0, avol0, n1, m) ↔ n0 + m ≤ # avol ∧ n1 + m ≤ # avol0 ∧ (∀ n. n < m → avol[n0 + n] = avol0[n1 + n]) subst: [m, n1, avol0, n0, avol] → [m0, 5 + n, avol0, 5 + n, avol] flags: <all paths>

Additional information

Used terms: m0, 5 + n, avol0, 5 + n, avol