1.42/0.67 YES 1.48/0.68 We consider the system theBenchmark. 1.48/0.68 1.48/0.68 Alphabet: 1.48/0.68 1.48/0.68 app : [] --> arrAB -> A -> B 1.48/0.68 case : [] --> SAB -> (A -> C) -> (B -> C) -> C 1.48/0.68 inl : [] --> A -> SAB 1.48/0.68 inr : [] --> B -> SAB 1.48/0.68 lam : [] --> (A -> B) -> arrAB 1.48/0.68 pair : [] --> A -> B -> PAB 1.48/0.68 piA : [] --> PAB -> A 1.48/0.68 piB : [] --> PAB -> B 1.48/0.68 1.48/0.68 Rules: 1.48/0.68 1.48/0.68 app (lam (/\x.f x)) y => f y 1.48/0.68 lam (/\x.app y x) => y 1.48/0.68 piA (pair x y) => x 1.48/0.68 piB (pair x y) => y 1.48/0.68 pair (piA x) (piB x) => x 1.48/0.68 case (inl x) (/\y.f y) (/\z.g z) => f x 1.48/0.68 case (inr x) (/\y.f y) (/\z.g z) => g x 1.48/0.68 1.48/0.68 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. 1.48/0.68 1.48/0.68 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: 1.48/0.68 1.48/0.68 Alphabet: 1.48/0.68 1.48/0.68 app : [arrAB * A] --> B 1.48/0.68 case : [SAB * A -> C * B -> C] --> C 1.48/0.68 inl : [A] --> SAB 1.48/0.68 inr : [B] --> SAB 1.48/0.68 lam : [A -> B] --> arrAB 1.48/0.68 pair : [A * B] --> PAB 1.48/0.68 piA : [PAB] --> A 1.48/0.68 piB : [PAB] --> B 1.48/0.68 ~AP1 : [A -> B * A] --> B 1.48/0.68 ~AP2 : [A -> C * A] --> C 1.48/0.68 ~AP3 : [B -> C * B] --> C 1.48/0.68 1.48/0.68 Rules: 1.48/0.68 1.48/0.68 app(lam(/\x.~AP1(F, x)), X) => ~AP1(F, X) 1.48/0.68 lam(/\x.app(X, x)) => X 1.48/0.68 piA(pair(X, Y)) => X 1.48/0.68 piB(pair(X, Y)) => Y 1.48/0.68 pair(piA(X), piB(X)) => X 1.48/0.68 case(inl(X), /\x.~AP2(F, x), /\y.~AP3(G, y)) => ~AP2(F, X) 1.48/0.68 case(inr(X), /\x.~AP2(F, x), /\y.~AP3(G, y)) => ~AP3(G, X) 1.48/0.68 app(lam(/\x.app(X, x)), Y) => app(X, Y) 1.48/0.68 ~AP1(F, X) => F X 1.48/0.68 ~AP2(F, X) => F X 1.48/0.68 ~AP3(F, X) => F X 1.48/0.68 1.48/0.68 Symbols ~AP1, ~AP2, and ~AP3 are encodings for application that are only used in innocuous ways. We can simplify the program (without losing non-termination) by removing them. 1.48/0.68 1.48/0.68 Additionally, we can remove some (now-)redundant rules. This gives: 1.48/0.68 1.48/0.68 Alphabet: 1.48/0.68 1.48/0.68 app : [arrAB * A] --> B 1.48/0.68 case : [SAB * A -> C * B -> C] --> C 1.48/0.68 inl : [A] --> SAB 1.48/0.68 inr : [B] --> SAB 1.48/0.68 lam : [A -> B] --> arrAB 1.48/0.68 pair : [A * B] --> PAB 1.48/0.68 piA : [PAB] --> A 1.48/0.68 piB : [PAB] --> B 1.48/0.68 1.48/0.68 Rules: 1.48/0.68 1.48/0.68 app(lam(/\x.X(x)), Y) => X(Y) 1.48/0.68 lam(/\x.app(X, x)) => X 1.48/0.68 piA(pair(X, Y)) => X 1.48/0.68 piB(pair(X, Y)) => Y 1.48/0.68 pair(piA(X), piB(X)) => X 1.48/0.68 case(inl(X), /\x.Y(x), /\y.Z(y)) => Y(X) 1.48/0.68 case(inr(X), /\x.Y(x), /\y.Z(y)) => Z(X) 1.48/0.68 1.48/0.68 We observe that the rules contain a first-order subset: 1.48/0.68 1.48/0.68 piA(pair(X, Y)) => X 1.48/0.68 piB(pair(X, Y)) => Y 1.48/0.68 pair(piA(X), piB(X)) => X 1.48/0.68 1.48/0.68 Moreover, the system is finitely branching. Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is Ce-terminating when seen as a many-sorted first-order TRS. 1.48/0.68 1.48/0.68 According to the external first-order termination prover, this system is indeed Ce-terminating: 1.48/0.68 1.48/0.68 || proof of resources/system.trs 1.48/0.68 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.48/0.68 || 1.48/0.68 || 1.48/0.68 || Termination w.r.t. Q of the given QTRS could be proven: 1.48/0.68 || 1.48/0.68 || (0) QTRS 1.48/0.68 || (1) QTRSRRRProof [EQUIVALENT] 1.48/0.68 || (2) QTRS 1.48/0.68 || (3) RisEmptyProof [EQUIVALENT] 1.48/0.68 || (4) YES 1.48/0.68 || 1.48/0.68 || 1.48/0.68 || ---------------------------------------- 1.48/0.68 || 1.48/0.68 || (0) 1.48/0.68 || Obligation: 1.48/0.68 || Q restricted rewrite system: 1.48/0.68 || The TRS R consists of the following rules: 1.48/0.68 || 1.48/0.68 || piA(pair(%X, %Y)) -> %X 1.48/0.68 || piB(pair(%X, %Y)) -> %Y 1.48/0.68 || pair(piA(%X), piB(%X)) -> %X 1.48/0.68 || ~PAIR(%X, %Y) -> %X 1.48/0.68 || ~PAIR(%X, %Y) -> %Y 1.48/0.68 || 1.48/0.68 || Q is empty. 1.48/0.68 || 1.48/0.68 || ---------------------------------------- 1.48/0.68 || 1.48/0.68 || (1) QTRSRRRProof (EQUIVALENT) 1.48/0.68 || Used ordering: 1.48/0.68 || Polynomial interpretation [POLO]: 1.48/0.68 || 1.48/0.68 || POL(pair(x_1, x_2)) = 1 + x_1 + x_2 1.48/0.68 || POL(piA(x_1)) = 1 + 2*x_1 1.48/0.68 || POL(piB(x_1)) = 1 + x_1 1.48/0.68 || POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2 1.48/0.68 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 1.48/0.68 || 1.48/0.68 || piA(pair(%X, %Y)) -> %X 1.48/0.68 || piB(pair(%X, %Y)) -> %Y 1.48/0.68 || pair(piA(%X), piB(%X)) -> %X 1.48/0.68 || ~PAIR(%X, %Y) -> %X 1.48/0.68 || ~PAIR(%X, %Y) -> %Y 1.48/0.68 || 1.48/0.68 || 1.48/0.68 || 1.48/0.68 || 1.48/0.68 || ---------------------------------------- 1.48/0.68 || 1.48/0.68 || (2) 1.48/0.68 || Obligation: 1.48/0.68 || Q restricted rewrite system: 1.48/0.68 || R is empty. 1.48/0.68 || Q is empty. 1.48/0.68 || 1.48/0.68 || ---------------------------------------- 1.48/0.68 || 1.48/0.68 || (3) RisEmptyProof (EQUIVALENT) 1.48/0.68 || The TRS R is empty. Hence, termination is trivially proven. 1.48/0.68 || ---------------------------------------- 1.48/0.68 || 1.48/0.68 || (4) 1.48/0.68 || YES 1.48/0.68 || 1.48/0.68 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]). 1.48/0.68 1.48/0.68 We thus obtain the following dependency pair problem (P_0, R_0, static, all): 1.48/0.68 1.48/0.68 Dependency Pairs P_0: 1.48/0.68 1.48/0.68 1.48/0.68 Rules R_0: 1.48/0.68 1.48/0.68 app(lam(/\x.X(x)), Y) => X(Y) 1.48/0.68 lam(/\x.app(X, x)) => X 1.48/0.68 piA(pair(X, Y)) => X 1.48/0.68 piB(pair(X, Y)) => Y 1.48/0.68 pair(piA(X), piB(X)) => X 1.48/0.68 case(inl(X), /\x.Y(x), /\y.Z(y)) => Y(X) 1.48/0.68 case(inr(X), /\x.Y(x), /\y.Z(y)) => Z(X) 1.48/0.68 1.48/0.68 Thus, the original system is terminating if (P_0, R_0, static, all) is finite. 1.48/0.68 1.48/0.68 We consider the dependency pair problem (P_0, R_0, static, all). 1.48/0.68 1.48/0.68 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 1.48/0.68 1.48/0.68 1.48/0.68 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 1.48/0.68 1.48/0.68 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 1.48/0.68 1.48/0.68 1.48/0.68 +++ Citations +++ 1.48/0.68 1.48/0.68 [Kop11] C. Kop. Simplifying Algebraic Functional Systems. In Proceedings of CAI 2011, volume 6742 of LNCS. 201--215, Springer, 2011. 1.48/0.68 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 1.48/0.68 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 1.48/0.68 [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. 1.48/0.68 EOF