We consider the system h46. Alphabet: 0 : [] --> nat plus : [] --> nat -> nat -> nat rec : [] --> nat -> nat -> (nat -> nat -> nat) -> nat s : [] --> nat -> nat succ : [] --> nat -> nat -> nat Rules: rec 0 x (/\y./\z.f y z) => x rec (s x) y (/\z./\u.f z u) => f x (rec x y (/\v./\w.f v w)) succ x y => s y plus x y => rec x y (/\z./\u.succ z u) 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: 0 : [] --> nat plus : [nat * nat] --> nat rec : [nat * nat * nat -> nat -> nat] --> nat s : [nat] --> nat succ : [nat * nat] --> nat ~AP1 : [nat -> nat -> nat * nat] --> nat -> nat Rules: rec(0, X, /\x./\y.~AP1(F, x) y) => X rec(s(X), Y, /\x./\y.~AP1(F, x) y) => ~AP1(F, X) rec(X, Y, /\z./\u.~AP1(F, z) u) succ(X, Y) => s(Y) plus(X, Y) => rec(X, Y, /\x./\y.succ(x, y)) rec(0, X, /\x./\y.plus(x, y)) => X rec(0, X, /\x./\y.succ(x, y)) => X rec(s(X), Y, /\x./\y.plus(x, y)) => plus(X, rec(X, Y, /\z./\u.plus(z, u))) rec(s(X), Y, /\x./\y.succ(x, y)) => succ(X, rec(X, Y, /\z./\u.succ(z, u))) ~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: 0 : [] --> nat plus : [nat * nat] --> nat rec : [nat * nat * nat -> nat -> nat] --> nat s : [nat] --> nat succ : [nat * nat] --> nat Rules: rec(0, X, /\x./\y.Y(x, y)) => X rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) succ(X, Y) => s(Y) plus(X, Y) => rec(X, Y, /\x./\y.succ(x, y)) We observe that the rules contain a first-order subset: succ(X, Y) => s(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: || || succ(%X, %Y) -> s(%Y) || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(s(x_1)) = 1 + x_1 || POL(succ(x_1, x_2)) = 2 + x_1 + 2*x_2 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || succ(%X, %Y) -> s(%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]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] rec#(s(X), Y, /\x./\y.Z(x, y)) =#> rec#(X, Y, /\z./\u.Z(z, u)) {Z : 2} 1] plus#(X, Y) =#> rec#(X, Y, /\x./\y.succ(x, y)) 2] plus#(X, Y) =#> succ#(Z, U) Rules R_0: rec(0, X, /\x./\y.Y(x, y)) => X rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) succ(X, Y) => s(Y) plus(X, Y) => rec(X, Y, /\x./\y.succ(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: * 0 : 0 * 1 : 0 * 2 : This graph has the following strongly connected components: P_1: rec#(s(X), Y, /\x./\y.Z(x, y)) =#> rec#(X, Y, /\z./\u.Z(z, u)) {Z : 2} 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(rec#) = 1 Thus, we can orient the dependency pairs as follows: nu(rec#(s(X), Y, /\x./\y.Z(x, y))) = s(X) |> X = nu(rec#(X, Y, /\z./\u.Z(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.