2.12/1.99 YES 2.12/2.00 We consider the system theBenchmark. 2.12/2.00 2.12/2.00 Alphabet: 2.12/2.00 2.12/2.00 cons : [c * d] --> d 2.12/2.00 f : [a * a] --> a 2.12/2.00 false : [] --> b 2.12/2.00 filter : [c -> b * d] --> d 2.12/2.00 filter2 : [b * c -> b * c * d] --> d 2.12/2.00 g : [a] --> a 2.12/2.00 h : [a] --> a 2.12/2.00 map : [c -> c * d] --> d 2.12/2.00 nil : [] --> d 2.12/2.00 true : [] --> b 2.12/2.00 2.12/2.00 Rules: 2.12/2.00 2.12/2.00 g(h(g(x))) => g(x) 2.12/2.00 g(g(x)) => g(h(g(x))) 2.12/2.00 h(h(x)) => h(f(h(x), x)) 2.12/2.00 map(i, nil) => nil 2.12/2.00 map(i, cons(x, y)) => cons(i x, map(i, y)) 2.12/2.00 filter(i, nil) => nil 2.12/2.00 filter(i, cons(x, y)) => filter2(i x, i, x, y) 2.12/2.00 filter2(true, i, x, y) => cons(x, filter(i, y)) 2.12/2.00 filter2(false, i, x, y) => filter(i, y) 2.12/2.00 2.12/2.00 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 2.12/2.00 2.12/2.00 We observe that the rules contain a first-order subset: 2.12/2.00 2.12/2.00 g(h(g(X))) => g(X) 2.12/2.00 g(g(X)) => g(h(g(X))) 2.12/2.00 h(h(X)) => h(f(h(X), X)) 2.12/2.00 2.12/2.00 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. 2.12/2.00 2.12/2.00 According to the external first-order termination prover, this system is indeed Ce-terminating: 2.12/2.00 2.12/2.00 || proof of resources/system.trs 2.12/2.00 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 2.12/2.00 || 2.12/2.00 || 2.12/2.00 || Termination w.r.t. Q of the given QTRS could be proven: 2.12/2.00 || 2.12/2.00 || (0) QTRS 2.12/2.00 || (1) DependencyPairsProof [EQUIVALENT] 2.12/2.00 || (2) QDP 2.12/2.00 || (3) DependencyGraphProof [EQUIVALENT] 2.12/2.00 || (4) QDP 2.12/2.00 || (5) QDPOrderProof [EQUIVALENT] 2.12/2.00 || (6) QDP 2.12/2.00 || (7) PisEmptyProof [EQUIVALENT] 2.12/2.00 || (8) YES 2.12/2.00 || 2.12/2.00 || 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (0) 2.12/2.00 || Obligation: 2.12/2.00 || Q restricted rewrite system: 2.12/2.00 || The TRS R consists of the following rules: 2.12/2.00 || 2.12/2.00 || g(h(g(%X))) -> g(%X) 2.12/2.00 || g(g(%X)) -> g(h(g(%X))) 2.12/2.00 || h(h(%X)) -> h(f(h(%X), %X)) 2.12/2.00 || ~PAIR(%X, %Y) -> %X 2.12/2.00 || ~PAIR(%X, %Y) -> %Y 2.12/2.00 || 2.12/2.00 || Q is empty. 2.12/2.00 || 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (1) DependencyPairsProof (EQUIVALENT) 2.12/2.00 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (2) 2.12/2.00 || Obligation: 2.12/2.00 || Q DP problem: 2.12/2.00 || The TRS P consists of the following rules: 2.12/2.00 || 2.12/2.00 || G(g(%X)) -> G(h(g(%X))) 2.12/2.00 || G(g(%X)) -> H(g(%X)) 2.12/2.00 || H(h(%X)) -> H(f(h(%X), %X)) 2.12/2.00 || 2.12/2.00 || The TRS R consists of the following rules: 2.12/2.00 || 2.12/2.00 || g(h(g(%X))) -> g(%X) 2.12/2.00 || g(g(%X)) -> g(h(g(%X))) 2.12/2.00 || h(h(%X)) -> h(f(h(%X), %X)) 2.12/2.00 || ~PAIR(%X, %Y) -> %X 2.12/2.00 || ~PAIR(%X, %Y) -> %Y 2.12/2.00 || 2.12/2.00 || Q is empty. 2.12/2.00 || We have to consider all minimal (P,Q,R)-chains. 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (3) DependencyGraphProof (EQUIVALENT) 2.12/2.00 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (4) 2.12/2.00 || Obligation: 2.12/2.00 || Q DP problem: 2.12/2.00 || The TRS P consists of the following rules: 2.12/2.00 || 2.12/2.00 || G(g(%X)) -> G(h(g(%X))) 2.12/2.00 || 2.12/2.00 || The TRS R consists of the following rules: 2.12/2.00 || 2.12/2.00 || g(h(g(%X))) -> g(%X) 2.12/2.00 || g(g(%X)) -> g(h(g(%X))) 2.12/2.00 || h(h(%X)) -> h(f(h(%X), %X)) 2.12/2.00 || ~PAIR(%X, %Y) -> %X 2.12/2.00 || ~PAIR(%X, %Y) -> %Y 2.12/2.00 || 2.12/2.00 || Q is empty. 2.12/2.00 || We have to consider all minimal (P,Q,R)-chains. 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (5) QDPOrderProof (EQUIVALENT) 2.12/2.00 || We use the reduction pair processor [LPAR04,JAR06]. 2.12/2.00 || 2.12/2.00 || 2.12/2.00 || The following pairs can be oriented strictly and are deleted. 2.12/2.00 || 2.12/2.00 || G(g(%X)) -> G(h(g(%X))) 2.12/2.00 || The remaining pairs can at least be oriented weakly. 2.12/2.00 || Used ordering: Polynomial interpretation [POLO]: 2.12/2.00 || 2.12/2.00 || POL(G(x_1)) = x_1 2.12/2.00 || POL(f(x_1, x_2)) = 0 2.12/2.00 || POL(g(x_1)) = 1 + x_1 2.12/2.00 || POL(h(x_1)) = 0 2.12/2.00 || 2.12/2.00 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 2.12/2.00 || 2.12/2.00 || h(h(%X)) -> h(f(h(%X), %X)) 2.12/2.00 || 2.12/2.00 || 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (6) 2.12/2.00 || Obligation: 2.12/2.00 || Q DP problem: 2.12/2.00 || P is empty. 2.12/2.00 || The TRS R consists of the following rules: 2.12/2.00 || 2.12/2.00 || g(h(g(%X))) -> g(%X) 2.12/2.00 || g(g(%X)) -> g(h(g(%X))) 2.12/2.00 || h(h(%X)) -> h(f(h(%X), %X)) 2.12/2.00 || ~PAIR(%X, %Y) -> %X 2.12/2.00 || ~PAIR(%X, %Y) -> %Y 2.12/2.00 || 2.12/2.00 || Q is empty. 2.12/2.00 || We have to consider all minimal (P,Q,R)-chains. 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (7) PisEmptyProof (EQUIVALENT) 2.12/2.00 || The TRS P is empty. Hence, there is no (P,Q,R) chain. 2.12/2.00 || ---------------------------------------- 2.12/2.00 || 2.12/2.00 || (8) 2.12/2.00 || YES 2.12/2.00 || 2.12/2.00 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 [Kop13]). 2.12/2.00 2.12/2.00 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 2.12/2.00 2.12/2.00 Dependency Pairs P_0: 2.12/2.00 2.12/2.00 0] map#(F, cons(X, Y)) =#> map#(F, Y) 2.12/2.00 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.12/2.00 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 2.12/2.00 3] filter2#(false, F, X, Y) =#> filter#(F, Y) 2.12/2.00 2.12/2.00 Rules R_0: 2.12/2.00 2.12/2.00 g(h(g(X))) => g(X) 2.12/2.00 g(g(X)) => g(h(g(X))) 2.12/2.00 h(h(X)) => h(f(h(X), X)) 2.12/2.00 map(F, nil) => nil 2.12/2.00 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 2.12/2.00 filter(F, nil) => nil 2.12/2.00 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 2.12/2.00 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 2.12/2.00 filter2(false, F, X, Y) => filter(F, Y) 2.12/2.00 2.12/2.00 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 2.12/2.00 2.12/2.00 We consider the dependency pair problem (P_0, R_0, static, formative). 2.12/2.00 2.12/2.00 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.12/2.00 2.12/2.00 * 0 : 0 2.12/2.00 * 1 : 2, 3 2.12/2.00 * 2 : 1 2.12/2.00 * 3 : 1 2.12/2.00 2.12/2.00 This graph has the following strongly connected components: 2.12/2.00 2.12/2.00 P_1: 2.12/2.00 2.12/2.00 map#(F, cons(X, Y)) =#> map#(F, Y) 2.12/2.00 2.12/2.00 P_2: 2.12/2.00 2.12/2.00 filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2.12/2.00 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.12/2.00 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.12/2.00 2.12/2.00 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) and (P_2, R_0, m, f). 2.12/2.00 2.12/2.00 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. 2.12/2.00 2.12/2.00 We consider the dependency pair problem (P_2, R_0, static, formative). 2.12/2.00 2.12/2.00 We apply the subterm criterion with the following projection function: 2.12/2.00 2.12/2.00 nu(filter2#) = 4 2.12/2.00 nu(filter#) = 2 2.12/2.00 2.12/2.00 Thus, we can orient the dependency pairs as follows: 2.12/2.00 2.12/2.00 nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 2.12/2.00 nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.12/2.00 nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 2.12/2.00 2.12/2.00 By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by (P_3, R_0, static, f), where P_3 contains: 2.12/2.00 2.12/2.00 filter2#(true, F, X, Y) =#> filter#(F, Y) 2.12/2.00 filter2#(false, F, X, Y) =#> filter#(F, Y) 2.12/2.00 2.12/2.00 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite. 2.12/2.00 2.12/2.00 We consider the dependency pair problem (P_3, R_0, static, formative). 2.12/2.00 2.12/2.00 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 2.12/2.00 2.12/2.00 * 0 : 2.12/2.00 * 1 : 2.12/2.00 2.12/2.00 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 2.12/2.00 2.12/2.00 Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. 2.12/2.00 2.12/2.00 We consider the dependency pair problem (P_1, R_0, static, formative). 2.12/2.00 2.12/2.00 We apply the subterm criterion with the following projection function: 2.12/2.00 2.12/2.00 nu(map#) = 2 2.12/2.00 2.12/2.00 Thus, we can orient the dependency pairs as follows: 2.12/2.00 2.12/2.00 nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 2.12/2.00 2.12/2.00 By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 2.12/2.00 2.12/2.00 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 2.12/2.00 2.12/2.00 2.12/2.00 +++ Citations +++ 2.12/2.00 2.12/2.00 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 2.12/2.00 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 2.12/2.00 [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. 2.12/2.00 EOF