Verification of concurrent data structures is one of the most challenging tasks in software verification. The topic has received considerable attention over the course of the last decade. Nevertheless, human-driven techniques remain cumbersome and notoriously difficult while automated approaches suffer from limited applicability. The main obstacle for automation is the complexity of concurrent data structures. This is particularly true in the absence of garbage collection. The intricacy of lock-free memory management paired with the complexity of concurrent data structures makes automated verification prohibitive.In this work we present a method for verifying concurrent data structures and their memory management separately. We suggest two simpler verification tasks that imply the correctness of the data structure. The first task establishes an over-approximation of the reclamation behavior of the memory management. The second task exploits this over-approximation to verify the data structure without the need to consider the implementation of the memory management itself. To make the resulting verification tasks tractable for automated techniques, we establish a second result. We show that a verification tool needs to consider only executions where a single memory location is reused. We implemented our approach and were able to verify linearizability of Michael&Scott's queue and the DGLM queue for both hazard pointers and epoch-based reclamation. To the best of our knowledge, we are the first to verify such implementations fully automatically.Otherwise, this equality may not hold.Property (G5) holds by definition together with Property (P5). It remains to show Properties (G2) to (G4) and (G6).Ad Property (G2). First consider the case where we have q ∈ valid τ . By Auxiliary (A1) we also have q ∈ valid σ . We getThe second equality holds because m τ (q) = b. That is, we preserve b.data in m τ | valid τ and only need to update the mapping of p. Similarly, we getNow consider the case q valid τ . By Auxiliary (A1) we also have q valid σ . We getThe last equality holds because the update does not survive the restriction to a set which is guaranteed not to contain p. Similarly, we getWe now get:The first equality is by definition of restrictions, the second equality is due to Property (P2), the third one is by Auxiliary (A1), and the last is again by definition. This concludes the property.Ad Property (G3). Only the valuation of p is changed by both act and act ′ . So by Property (P3) it suffices to show m τ .act (p)=a ⇐⇒ m σ .act ′ (p)=a. This follows easily:where the first and last equivalence hold due to the updates up and up ′ and the second equality holds by Property (P3).Ad Property (G4). Let c ∈ m τ .act (valid τ .act ). We have c ∈ m τ (valid τ ). To see this, consider some pexp ∈ valid τ .act such that m τ .act (pexp) = c. If pexp p, then m τ .act (pexp) = m τ (pexp) and pexp ∈ valid τ by definition. So c ∈ m τ (valid τ ) follows immediately. Otherwise, in the case of pexp ≡ p, we have m τ .act (p) = m τ (q). ...