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Proof step 2 from proof rangeeq-till-leb5-fwd (previous step)

m < 5, range=(avol1, 0, avol0, 0, 5 + n) ⊦ avol1[m] = avol0[m]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ range=(avol, n0, avol0, n1, m) ↔ n0 + m ≤ # avol ∧ n1 + m ≤ # avol0 ∧ (∀ n. n < m → avol[n0 + n] = avol0[n1 + n])
Proof step 4:
m < 5, 0 + 5 + n ≤ # avol1, 0 + 5 + n ≤ # avol0, ∀ n2. n2 < 5 + n → avol1[0 + n2] = avol0[0 + n2] ⊦ avol1[m] = avol0[m]

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] → [5 + n, 0, avol0, 0, avol1] flags: <all paths>

Additional information

Used terms: 5 + n, 0, avol0, 0, avol1