We consider the system h61. Alphabet: app : [] --> list -> list -> list cons : [] --> nat -> list -> list foldl : [] --> (list -> nat -> list) -> list -> list -> list iconsc : [] --> list -> nat -> list nil : [] --> list reverse : [] --> list -> list reverse1 : [] --> list -> list Rules: app nil x => x app (cons x y) z => cons x (app y z) foldl (/\x./\y.f x y) z nil => z foldl (/\x./\y.f x y) z (cons u v) => foldl (/\w./\x'.f w x') (f z u) v iconsc x y => cons y x reverse x => foldl (/\y./\z.iconsc y z) nil x reverse1 x => foldl (/\y./\z.app (cons z nil) y) nil x 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. 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: Alphabet: app : [list * list] --> list cons : [nat * list] --> list foldl : [list -> nat -> list * list * list] --> list iconsc : [list * nat] --> list nil : [] --> list reverse : [list] --> list reverse1 : [list] --> list ~AP1 : [list -> nat -> list * list] --> nat -> list Rules: app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) foldl(/\x./\y.~AP1(F, x) y, X, nil) => X foldl(/\x./\y.~AP1(F, x) y, X, cons(Y, Z)) => foldl(/\z./\u.~AP1(F, z) u, ~AP1(F, X) Y, Z) iconsc(X, Y) => cons(Y, X) reverse(X) => foldl(/\x./\y.iconsc(x, y), nil, X) reverse1(X) => foldl(/\x./\y.app(cons(y, nil), x), nil, X) foldl(/\x./\y.iconsc(x, y), X, nil) => X foldl(/\x./\y.iconsc(x, y), X, cons(Y, Z)) => foldl(/\z./\u.iconsc(z, u), iconsc(X, Y), Z) ~AP1(F, X) => F X 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: Alphabet: app : [list * list] --> list cons : [nat * list] --> list foldl : [list -> nat -> list * list * list] --> list iconsc : [list * nat] --> list nil : [] --> list reverse : [list] --> list reverse1 : [list] --> list Rules: app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) foldl(/\x./\y.X(x, y), Y, nil) => Y foldl(/\x./\y.X(x, y), Y, cons(Z, U)) => foldl(/\z./\u.X(z, u), X(Y, Z), U) iconsc(X, Y) => cons(Y, X) reverse(X) => foldl(/\x./\y.iconsc(x, y), nil, X) reverse1(X) => foldl(/\x./\y.app(cons(y, nil), x), nil, X) We observe that the rules contain a first-order subset: app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) iconsc(X, Y) => cons(Y, X) 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: || || app(nil, %X) -> %X || app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z)) || iconsc(%X, %Y) -> cons(%Y, %X) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(app(x_1, x_2)) = 2 + 2*x_1 + x_2 || POL(cons(x_1, x_2)) = 1 + x_1 + x_2 || POL(iconsc(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 || POL(nil) = 1 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || app(nil, %X) -> %X || app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z)) || iconsc(%X, %Y) -> cons(%Y, %X) || || || || || ---------------------------------------- || || (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]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> foldl#(/\z./\u.X(z, u), X(Y, Z), U) 1] reverse#(X) =#> foldl#(/\x./\y.iconsc(x, y), nil, X) 2] reverse#(X) =#> iconsc#(Y, Z) 3] reverse1#(X) =#> foldl#(/\x./\y.app(cons(y, nil), x), nil, X) 4] reverse1#(X) =#> app#(cons(Y, nil), Z) Rules R_0: app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) foldl(/\x./\y.X(x, y), Y, nil) => Y foldl(/\x./\y.X(x, y), Y, cons(Z, U)) => foldl(/\z./\u.X(z, u), X(Y, Z), U) iconsc(X, Y) => cons(Y, X) reverse(X) => foldl(/\x./\y.iconsc(x, y), nil, X) reverse1(X) => foldl(/\x./\y.app(cons(y, nil), x), nil, X) 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: * 0 : 0 * 1 : 0 * 2 : * 3 : 0 * 4 : This graph has the following strongly connected components: P_1: foldl#(/\x./\y.X(x, y), Y, cons(Z, U)) =#> foldl#(/\z./\u.X(z, u), X(Y, Z), U) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f). Thus, the original system is terminating if (P_1, R_0, computable, formative) is finite. We consider the dependency pair problem (P_1, R_0, computable, formative). We apply the subterm criterion with the following projection function: nu(foldl#) = 3 Thus, we can orient the dependency pairs as follows: nu(foldl#(/\x./\y.X(x, y), Y, cons(Z, U))) = cons(Z, U) |> U = nu(foldl#(/\z./\u.X(z, u), X(Y, Z), U)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 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. [Kop11] C. Kop. Simplifying Algebraic Functional Systems. In Proceedings of CAI 2011, volume 6742 of LNCS. 201--215, Springer, 2011. [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.