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Proof step 57 from proof range-eq-splice-less (previous step)

n3 < m0, n3 < m1, # ar0 < n0 + m1, # ar0 < n0 + m0, n2 + m1 ≤ # ar2, n + m0 ≤ # ar, m1 ≤ m0, n1 ≤ n0, ∀ n3. n3 < m1 → ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m1), n0, m1)[n1 + n3] ⊦ ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m0), n0, m0)[n1 + n3]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 58:
⊦ m ≤ n ↔ (∃ n0. m + n0 = n)
Proof step 59:
# ar0 < n0 + m0, # ar0 < n0 + m1, n3 < m1, n3 < m0, m1 ≤ m0, n + m0 ≤ # ar, n2 + m1 ≤ # ar2, ∀ n3. n3 < m1 → ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m1), n0, m1)[n1 + n3], ∃ n2. n1 + n2 = n0 ⊦ ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m0), n0, m0)[n1 + n3]

Rule argument

spec: nat name: le-addex seq: ⊦ m ≤ n ↔ (∃ n0. m + n0 = n) subst: [n, m] → [n0, n1] flags: <all paths>

Additional information

Used terms: n0, n1