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Proof step 2 from proof copy-coalesce (previous step)

m0 ≤ # ar1, m0 + n ≤ # ar1 ⊦ copy(copy(ar0, n0, ar1, 0, # ar1), n, ar2, n2, m0) = copy(ar0, n0 + n, ar2, n2, m0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 4:
m0 ≤ # ar1, m0 + n ≤ # ar1 ⊦ # copy(copy(ar0, n0, ar1, 0, # ar1), n, ar2, n2, m0) = # copy(ar0, n0 + n, ar2, n2, m0) ∧ (∀ n1. n1 < # copy(copy(ar0, n0, ar1, 0, # ar1), n, ar2, n2, m0) → copy(copy(ar0, n0, ar1, 0, # ar1), n, ar2, n2, m0)[n1] = copy(ar0, n0 + n, ar2, n2, m0)[n1])

Rule argument

spec: array inst: name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [copy(ar0, n0 + n, ar2, n2, m0), copy(copy(ar0, n0, ar1, 0, # ar1), n, ar2, n2, m0)] flags: rewrite

Additional information

Used terms: copy(ar0, n0 + n, ar2, n2, m0), copy(copy(ar0, n0, ar1, 0, # ar1), n, ar2, n2, m0)