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Proof step 8 from proof lebs-range-eq (previous step)

n0 + n < m + m2, # avol0 < m + m2, range=(avol0, m, avol1, m, m2), m ≤ n0, # avol0 ≤ n0 + n, ∀ avol0, m, avol1, m2, n0. range=(avol0, m, avol1, m, m2) ∧ n0 + n ≤ m + m2 ∧ m ≤ n0 → lebs(avol0, n0, n) = lebs(avol1, n0, n) ⊦ lebs(avol0, n0, n + 1) = lebs(avol1, n0, n + 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ range=(avol, n0, avol0, n1, m) ↔ n0 + m ≤ # avol ∧ n1 + m ≤ # avol0 ∧ (∀ n. n < m → avol[n0 + n] = avol0[n1 + n])

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] → [m2, m, avol1, m, avol0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: m2, m, avol1, m, avol0