We consider the system Applicative_05__ReverseLastInit. Alphabet: compose : [a -> a * a -> a] --> a -> a cons : [a * a] --> a hd : [] --> a -> a init : [] --> a -> a last : [] --> a -> a nil : [] --> a reverse : [] --> a -> a reverse2 : [a * a] --> a tl : [] --> a -> a Rules: compose(f, g) x => g (f x) reverse x => reverse2(x, nil) reverse2(nil, x) => x reverse2(cons(x, y), z) => reverse2(y, cons(x, z)) hd cons(x, y) => x tl cons(x, y) => y last => compose(hd, reverse) init => compose(reverse, compose(tl, reverse)) 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: reverse(X) => reverse2(X, nil) reverse2(nil, X) => X reverse2(cons(X, Y), Z) => reverse2(Y, cons(X, Z)) hd(cons(X, Y)) => X tl(cons(X, Y)) => 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: || || reverse(%X) -> reverse2(%X, nil) || reverse2(nil, %X) -> %X || reverse2(cons(%X, %Y), %Z) -> reverse2(%Y, cons(%X, %Z)) || hd(cons(%X, %Y)) -> %X || tl(cons(%X, %Y)) -> %Y || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(cons(x_1, x_2)) = 1 + x_1 + x_2 || POL(hd(x_1)) = 2 + x_1 || POL(nil) = 2 || POL(reverse(x_1)) = 2 + 2*x_1 || POL(reverse2(x_1, x_2)) = 2*x_1 + x_2 || POL(tl(x_1)) = 2 + x_1 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || reverse2(nil, %X) -> %X || reverse2(cons(%X, %Y), %Z) -> reverse2(%Y, cons(%X, %Z)) || hd(cons(%X, %Y)) -> %X || tl(cons(%X, %Y)) -> %Y || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || reverse(%X) -> reverse2(%X, nil) || || Q is empty. || || ---------------------------------------- || || (3) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(nil) = 0 || POL(reverse(x_1)) = 2 + 2*x_1 || POL(reverse2(x_1, x_2)) = 1 + 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: || || reverse(%X) -> reverse2(%X, nil) || || || || || ---------------------------------------- || || (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 and accessible arguments in [FuhKop19]). In order to do so, we start by eta-expanding the system, which gives: compose(F, G, X) => G (F X) reverse(X) => reverse2(X, nil) reverse2(nil, X) => X reverse2(cons(X, Y), Z) => reverse2(Y, cons(X, Z)) hd(cons(X, Y)) => X tl(cons(X, Y)) => Y last(X) => compose(/\x.hd(x), /\y.reverse(y), X) init(X) => compose(/\x.reverse(x), /\y.compose(/\z.tl(z), /\u.reverse(u), y), X) We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] last#(X) =#> compose#(/\x.hd(x), /\y.reverse(y), X) 1] last#(X) =#> hd#(Y) 2] last#(X) =#> reverse#(Y) 3] init#(X) =#> compose#(/\x.reverse(x), /\y.compose(/\z.tl(z), /\u.reverse(u), y), X) 4] init#(X) =#> reverse#(Y) 5] init#(X) =#> compose#(/\x.tl(x), /\y.reverse(y), Y) 6] init#(X) =#> tl#(Y) 7] init#(X) =#> reverse#(Y) Rules R_0: compose(F, G, X) => G (F X) reverse(X) => reverse2(X, nil) reverse2(nil, X) => X reverse2(cons(X, Y), Z) => reverse2(Y, cons(X, Z)) hd(cons(X, Y)) => X tl(cons(X, Y)) => Y last(X) => compose(/\x.hd(x), /\y.reverse(y), X) init(X) => compose(/\x.reverse(x), /\y.compose(/\z.tl(z), /\u.reverse(u), y), 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 : * 4 : * 5 : * 6 : * 7 : 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.