We consider the system h07. Alphabet: 0 : [] --> N even : [] --> N -> N -> B false : [] --> B g : [] --> N -> B h : [] --> N -> (N -> B) -> N -> B not : [] --> B -> B rec : [] --> (N -> (N -> B) -> N -> B) -> B -> N -> B true : [] --> B Rules: rec f (i 0) => i g x => true h x f y => not (f y) not true => false not false => true even x y => rec h (g x) y 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 : [] --> N even : [N] --> N -> B false : [] --> B g : [] --> N -> B h : [] --> N -> (N -> B) -> N -> B not : [B] --> B rec : [N -> (N -> B) -> N -> B * B] --> N -> B true : [] --> B ~AP1 : [N -> B * N] --> B Rules: rec(F, ~AP1(G, 0)) => G g X => true h X F Y => not(~AP1(F, Y)) not(true) => false not(false) => true even(X) Y => ~AP1(rec(h, g X), Y) rec(F, even(X) 0) => even(X) rec(F, g 0) => g rec(F, h X G 0) => h X G ~AP1(F, X) => F X We observe that the rules contain a first-order subset: g(X) => true not(true) => false not(false) => true 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. According to the external first-order termination prover, this system is indeed Ce-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: || || g(%X) -> true || not(true) -> false || not(false) -> true || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(false) = 2 || POL(g(x_1)) = 2 + x_1 || POL(not(x_1)) = 2 + 2*x_1 || POL(true) = 1 || POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || g(%X) -> true || not(true) -> false || not(false) -> true || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %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: rec(F, ~AP1(G, 0), X) => G X g(X) => true h(X, F, Y) => not(~AP1(F, Y)) not(true) => false not(false) => true even(X, Y) => ~AP1(/\x.rec(/\y./\f./\z.h(y, /\u.f u, z), g(X), x), Y) rec(F, even(X, 0), Y) => even(X, Y) rec(F, g(0), X) => g(X) rec(F, h(X, G, 0), Y) => h(X, G, Y) ~AP1(F, X) => F X We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] h#(X, F, Y) =#> not#(~AP1(F, Y)) 1] h#(X, F, Y) =#> ~AP1#(F, Y) 2] even#(X, Y) =#> ~AP1#(/\x.rec(/\y./\f./\z.h(y, /\u.f u, z), g(X), x), Y) 3] even#(X, Y) =#> rec#(/\x./\f./\y.h(x, /\z.f z, y), g(X), Z) 4] even#(X, Y) =#> h#(Z, /\x.F x, U) 5] even#(X, Y) =#> g#(X) 6] rec#(F, even(X, 0), Y) =#> even#(X, Y) 7] rec#(F, g(0), X) =#> g#(X) 8] rec#(F, h(X, G, 0), Y) =#> h#(X, G, Y) Rules R_0: rec(F, ~AP1(G, 0), X) => G X g(X) => true h(X, F, Y) => not(~AP1(F, Y)) not(true) => false not(false) => true even(X, Y) => ~AP1(/\x.rec(/\y./\f./\z.h(y, /\u.f u, z), g(X), x), Y) rec(F, even(X, 0), Y) => even(X, Y) rec(F, g(0), X) => g(X) rec(F, h(X, G, 0), Y) => h(X, G, Y) ~AP1(F, X) => F 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 : * 1 : * 2 : * 3 : 7 * 4 : 0, 1 * 5 : * 6 : 2, 3, 4, 5 * 7 : * 8 : 0, 1 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. [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.