\begin{circuslaw}{distr-prefix-par-1}
\begin{circusaction*}
  c~?~x \then (A_1 \lpar ns_1 | cs | ns_2 \rpar A_2) ~ \equiv ~ (c~?~x \then A_1 \lpar ns_1 | cs \cup \{c\} | ns_2 \rpar c~?~x \then A_2)
  \also
  \provided \; c \notin \usedC(A_1) \; \provand \; c \notin \usedC(A_2)
\end{circusaction*}
\end{circuslaw}
