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Proof step 4 from proof copy-append (previous step)

n0 + m1 ≤ # ar1 ⊦ copy(ar0 + ar1, n0 + # ar0, ar2, m0, m1) = copy(ar1, n0, ar2, m0, m1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 6:
n0 + m1 ≤ # ar1 ⊦ # copy(ar0 + ar1, n0 + # ar0, ar2, m0, m1) = # copy(ar1, n0, ar2, m0, m1) ∧ (∀ n. n < # copy(ar0 + ar1, n0 + # ar0, ar2, m0, m1) → copy(ar0 + ar1, n0 + # ar0, ar2, m0, m1)[n] = copy(ar1, n0, ar2, m0, m1)[n])

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [copy(ar1, n0, ar2, m0, m1), copy(ar0 + ar1, n0 + # ar0, ar2, m0, m1)] flags: <all paths>

Additional information

Used terms: copy(ar1, n0, ar2, m0, m1), copy(ar0 + ar1, n0 + # ar0, ar2, m0, m1)