We consider the system uncurry. Alphabet: f : [] --> a -> b -> c f1 : [a] --> b -> c f2 : [a * b] --> c Rules: f x => f1(x) f1(x) y => f2(x, y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We observe that the rules contain a first-order subset: f(X, Y) => f1(X, Y) f1(X, Y) => f2(X, Y) Moreover, the system is orthogonal. 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 terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed terminating: || proof of resources/system.trs || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 || || || Termination w.r.t. Q of the given QTRS could be proven: || || (0) QTRS || (1) QTRSRRRProof [EQUIVALENT] || (2) QTRS || (3) QTRSRRRProof [EQUIVALENT] || (4) QTRS || (5) RisEmptyProof [EQUIVALENT] || (6) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || f(%X, %Y) -> f1(%X, %Y) || f1(%X, %Y) -> f2(%X, %Y) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(f(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 || POL(f1(x_1, x_2)) = 2 + x_1 + 2*x_2 || POL(f2(x_1, x_2)) = 1 + x_1 + x_2 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || f1(%X, %Y) -> f2(%X, %Y) || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || f(%X, %Y) -> f1(%X, %Y) || || Q is empty. || || ---------------------------------------- || || (3) QTRSRRRProof (EQUIVALENT) || Used ordering: || Knuth-Bendix order [KBO] with precedence:f_2 > f1_2 || || and weight map: || || f_2=0 || f1_2=0 || || The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || f(%X, %Y) -> f1(%X, %Y) || || || || || ---------------------------------------- || || (4) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (5) RisEmptyProof (EQUIVALENT) || The TRS R is empty. Hence, termination is trivially proven. || ---------------------------------------- || || (6) || YES || We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). In order to do so, we start by eta-expanding the system, which gives: f(X, Y) => f1(X, Y) f1(X, Y) => f2(X, Y) We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: Rules R_0: f(X, Y) => f1(X, Y) f1(X, Y) => f2(X, Y) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [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.