The C11 and C++11 Concurrency Model

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• scC(client(X))∪scL(lib(X)) is acyclic.
Proof Sketch It is straightforward that client(X) and lib(X) satisfy Noninterf. Furthermore,
any cycle in scC(client(X))∪scL(lib(X)) is also a cycle the original execution,
and this contradicts the consistency of SC.
To check 8.1 (8.2 is similar), we need to show that the client projection of the happensbefore
edges in X match the happens-before relation created by the most general-client
execution of the client extended by the library core guarantee. Note that the MGC
happens-before is certainly a subset, so it remains to show that any happens-before edge
in client(X) is also an edge in an execution of the most general client. Consider an
arbitrary edge between actions u and v. The happens-before edge is made up of a path
in sb and sw. All interface actions on the same thread are related by sequenced before,
so we can pick out all interface actions in this path. Synchronisation is only ever created
between actions at the same location, so non-interference implies that there is never a
synchronises-with edge between the two components. Together that implies that any
segment of library actions in the path of edges from u to v starts with a call action and
ends with a return action, and is covered by the history, as required.
It is clear that the most general client will generate pre-executions that cover the
projections in each case above. Any subset of the actions of an execution together with
the relations restricted to that set will satisfy most of the conjuncts of the consistency
predicate, with those conjuncts that deal with locks, and read values implied by noninterference.
Because X satisfies non-interference, any safety violation must either be a data race,
an indeterminate read or an instance of bad mutex use. In the racy case, again because
of non-interference, both racy actions must be in the same component, as required, The
mutex case is similar. Indeterminate reads are a single action fault, so they reside in one
component or the other.
Lemma 23 (Composition). Consider
X ∈ C, hbL(core(Y))I;
Y ∈ L 2 , hbC(core(X))I 2 ,
such that
• A(X)∩A(Y) = interf(X) = interf(Y) and otherwise action and atomic section identifiers
in A(X) and A(Y) are different;
• interf(sb(X)) = interf(sb(Y));
• scC(X)∪scL(Y) is acyclic; and
• X and Y satisfy Noninterf.