We consider the system AotoYamada_05__005. Alphabet: 0 : [] --> a add : [] --> a -> a -> a curry : [a -> a -> a] --> a -> a -> a plus : [] --> a -> a -> a s : [a] --> a Rules: plus 0 x => x plus s(x) y => s(plus x y) curry(f) x y => f x y add => curry(plus) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). Symbol curry is an encoding for application that is only used in innocuous ways. We can simplify the program (without losing non-termination) by removing it. This gives: Alphabet: 0 : [] --> a add : [] --> a -> a -> a plus : [] --> a -> a -> a s : [a] --> a Rules: plus 0 X => X plus s(X) Y => s(plus X Y) add => plus We observe that the rules contain a first-order subset: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) add(X, Y) => plus(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) RisEmptyProof [EQUIVALENT] || (4) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || add(%X, %Y) -> plus(%X, %Y) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Knuth-Bendix order [KBO] with precedence:add_2 > plus_2 > s_1 > 0 || || and weight map: || || 0=1 || s_1=1 || plus_2=0 || add_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: || || plus(0, %X) -> %X || plus(s(%X), %Y) -> s(plus(%X, %Y)) || add(%X, %Y) -> plus(%X, %Y) || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (3) RisEmptyProof (EQUIVALENT) || The TRS R is empty. Hence, termination is trivially proven. || ---------------------------------------- || || (4) || 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 and accessible arguments in [FuhKop19]). In order to do so, we start by eta-expanding the system, which gives: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) add(X, Y) => plus(X, Y) We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: Rules R_0: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) add(X, Y) => plus(X, Y) Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite. We consider the dependency pair problem (P_0, R_0, computable, 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 +++ [FuhKop19] C. Fuhs, and C. Kop. A static higher-order dependency pair framework. In Proceedings of ESOP 2019, 2019. [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.