(* Diffie-Hellman *)

data g/0.
fun exp/2.

equation exp(exp(g,x),y) = exp(exp(g,y),x).

free c.

process
	new a1; new a2; new a3;
	out(c, (exp(g,a1), exp(g,a2), choice[exp(exp(g,a1),a2), exp(g,a3)]))

(* EXPECTPV
RESULT Observational equivalence is true.
0.036s (user 0.032s + system 0.004s), max rss 8804K
END *)
