0.00/0.02 YES 0.00/0.02 We consider the system theBenchmark. 0.00/0.02 0.00/0.02 Alphabet: 0.00/0.02 0.00/0.02 app : [] --> arrab -> a -> b 0.00/0.02 lam : [] --> (a -> b) -> arrab 0.00/0.02 0.00/0.02 Rules: 0.00/0.02 0.00/0.02 app (lam (/\x.f x)) y => f y 0.00/0.02 lam (/\x.app y x) => y 0.00/0.02 0.00/0.02 Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. 0.00/0.02 0.00/0.02 We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: 0.00/0.02 0.00/0.02 Alphabet: 0.00/0.02 0.00/0.02 app : [arrab * a] --> b 0.00/0.02 lam : [a -> b] --> arrab 0.00/0.02 ~AP1 : [a -> b * a] --> b 0.00/0.02 0.00/0.02 Rules: 0.00/0.02 0.00/0.02 app(lam(/\x.~AP1(F, x)), X) => ~AP1(F, X) 0.00/0.02 lam(/\x.app(X, x)) => X 0.00/0.02 app(lam(/\x.app(X, x)), Y) => app(X, Y) 0.00/0.02 ~AP1(F, X) => F X 0.00/0.02 0.00/0.02 Symbol ~AP1 is an encoding for application that is only used in innocuous ways. We can simplify the program (without losing non-termination) by removing it. Additionally, we can remove some (now-)redundant rules. This gives: 0.00/0.02 0.00/0.02 Alphabet: 0.00/0.02 0.00/0.02 app : [arrab * a] --> b 0.00/0.02 lam : [a -> b] --> arrab 0.00/0.02 0.00/0.02 Rules: 0.00/0.02 0.00/0.02 app(lam(/\x.X(x)), Y) => X(Y) 0.00/0.02 lam(/\x.app(X, x)) => X 0.00/0.02 0.00/0.02 We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]). 0.00/0.02 0.00/0.02 We thus obtain the following dependency pair problem (P_0, R_0, static, all): 0.00/0.02 0.00/0.02 Dependency Pairs P_0: 0.00/0.02 0.00/0.02 0.00/0.02 Rules R_0: 0.00/0.02 0.00/0.02 app(lam(/\x.X(x)), Y) => X(Y) 0.00/0.02 lam(/\x.app(X, x)) => X 0.00/0.02 0.00/0.02 Thus, the original system is terminating if (P_0, R_0, static, all) is finite. 0.00/0.02 0.00/0.02 We consider the dependency pair problem (P_0, R_0, static, all). 0.00/0.02 0.00/0.02 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 0.00/0.02 0.00/0.02 0.00/0.02 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 0.00/0.02 0.00/0.02 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 0.00/0.02 0.00/0.02 0.00/0.02 +++ Citations +++ 0.00/0.02 0.00/0.02 [Kop11] C. Kop. Simplifying Algebraic Functional Systems. In Proceedings of CAI 2011, volume 6742 of LNCS. 201--215, Springer, 2011. 0.00/0.02 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.02 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 0.00/0.02 [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009. 0.00/0.02 EOF