1.62/0.82 YES 1.62/0.82 We consider the system theBenchmark. 1.62/0.82 1.62/0.82 Alphabet: 1.62/0.82 1.62/0.82 cons : [c * d] --> d 1.62/0.82 f : [a] --> a 1.62/0.82 false : [] --> b 1.62/0.82 filter : [c -> b * d] --> d 1.62/0.82 filter2 : [b * c -> b * c * d] --> d 1.62/0.82 g : [a] --> a 1.62/0.82 h : [a] --> a 1.62/0.82 map : [c -> c * d] --> d 1.62/0.82 nil : [] --> d 1.62/0.82 true : [] --> b 1.62/0.82 1.62/0.82 Rules: 1.62/0.82 1.62/0.82 f(g(x)) => g(f(f(x))) 1.62/0.82 f(h(x)) => h(g(x)) 1.62/0.82 map(i, nil) => nil 1.62/0.82 map(i, cons(x, y)) => cons(i x, map(i, y)) 1.62/0.82 filter(i, nil) => nil 1.62/0.82 filter(i, cons(x, y)) => filter2(i x, i, x, y) 1.62/0.82 filter2(true, i, x, y) => cons(x, filter(i, y)) 1.62/0.82 filter2(false, i, x, y) => filter(i, y) 1.62/0.82 1.62/0.82 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 1.62/0.82 1.62/0.82 We use rule removal, following [Kop12, Theorem 2.23]. 1.62/0.82 1.62/0.82 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 1.62/0.82 1.62/0.82 f(g(X)) >? g(f(f(X))) 1.62/0.82 f(h(X)) >? h(g(X)) 1.62/0.82 map(F, nil) >? nil 1.62/0.82 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 1.62/0.82 filter(F, nil) >? nil 1.62/0.82 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 1.62/0.82 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 1.62/0.82 filter2(false, F, X, Y) >? filter(F, Y) 1.62/0.82 1.62/0.82 We orient these requirements with a polynomial interpretation in the natural numbers. 1.62/0.82 1.62/0.82 The following interpretation satisfies the requirements: 1.62/0.82 1.62/0.82 cons = \y0y1.1 + y0 + y1 1.62/0.82 f = \y0.y0 1.62/0.82 false = 3 1.62/0.82 filter = \G0y1.2 + 2y1 + G0(0) + 2y1G0(y1) 1.62/0.82 filter2 = \y0G1y2y3.1 + y0 + y2 + 2y3 + G1(0) + 2y3G1(y3) 1.62/0.82 g = \y0.y0 1.62/0.82 h = \y0.y0 1.62/0.82 map = \G0y1.2 + 3y1 + 2y1G0(y1) + 2G0(y1) 1.62/0.82 nil = 0 1.62/0.82 true = 3 1.62/0.82 1.62/0.82 Using this interpretation, the requirements translate to: 1.62/0.82 1.62/0.82 [[f(g(_x0))]] = x0 >= x0 = [[g(f(f(_x0)))]] 1.62/0.82 [[f(h(_x0))]] = x0 >= x0 = [[h(g(_x0))]] 1.62/0.82 [[map(_F0, nil)]] = 2 + 2F0(0) > 0 = [[nil]] 1.62/0.82 [[map(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 4F0(1 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + 2x2F0(x2) + 2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 1.62/0.82 [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 1.62/0.82 [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 1 + 2x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 1.62/0.82 [[filter2(true, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 1.62/0.82 [[filter2(false, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2 + 2x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 1.62/0.82 1.62/0.82 We can thus remove the following rules: 1.62/0.82 1.62/0.82 map(F, nil) => nil 1.62/0.82 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 1.62/0.82 filter(F, nil) => nil 1.62/0.82 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 1.62/0.82 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 1.62/0.82 filter2(false, F, X, Y) => filter(F, Y) 1.62/0.82 1.62/0.82 We observe that the rules contain a first-order subset: 1.62/0.82 1.62/0.82 f(g(X)) => g(f(f(X))) 1.62/0.82 f(h(X)) => h(g(X)) 1.62/0.82 1.62/0.82 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. 1.62/0.82 1.62/0.82 According to the external first-order termination prover, this system is indeed terminating: 1.62/0.82 1.62/0.82 || proof of resources/system.trs 1.62/0.82 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.62/0.82 || 1.62/0.82 || 1.62/0.82 || Termination w.r.t. Q of the given QTRS could be proven: 1.62/0.82 || 1.62/0.82 || (0) QTRS 1.62/0.82 || (1) QTRS Reverse [EQUIVALENT] 1.62/0.82 || (2) QTRS 1.62/0.82 || (3) RFCMatchBoundsTRSProof [EQUIVALENT] 1.62/0.82 || (4) YES 1.62/0.82 || 1.62/0.82 || 1.62/0.82 || ---------------------------------------- 1.62/0.82 || 1.62/0.82 || (0) 1.62/0.82 || Obligation: 1.62/0.82 || Q restricted rewrite system: 1.62/0.82 || The TRS R consists of the following rules: 1.62/0.82 || 1.62/0.82 || f(g(%X)) -> g(f(f(%X))) 1.62/0.82 || f(h(%X)) -> h(g(%X)) 1.62/0.82 || 1.62/0.82 || Q is empty. 1.62/0.82 || 1.62/0.82 || ---------------------------------------- 1.62/0.82 || 1.62/0.82 || (1) QTRS Reverse (EQUIVALENT) 1.62/0.82 || We applied the QTRS Reverse Processor [REVERSE]. 1.62/0.82 || ---------------------------------------- 1.62/0.82 || 1.62/0.82 || (2) 1.62/0.82 || Obligation: 1.62/0.82 || Q restricted rewrite system: 1.62/0.82 || The TRS R consists of the following rules: 1.62/0.82 || 1.62/0.82 || g(f(x)) -> f(f(g(x))) 1.62/0.82 || h(f(x)) -> g(h(x)) 1.62/0.82 || 1.62/0.82 || Q is empty. 1.62/0.82 || 1.62/0.82 || ---------------------------------------- 1.62/0.82 || 1.62/0.82 || (3) RFCMatchBoundsTRSProof (EQUIVALENT) 1.62/0.82 || Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R. 1.62/0.82 || The following rules were used to construct the certificate: 1.62/0.82 || 1.62/0.82 || g(f(x)) -> f(f(g(x))) 1.62/0.82 || h(f(x)) -> g(h(x)) 1.62/0.82 || 1.62/0.82 || The certificate found is represented by the following graph. 1.62/0.82 || The certificate consists of the following enumerated nodes: 1.62/0.82 || 3, 4, 10, 11, 19, 20, 21, 22 1.62/0.82 || 1.62/0.82 || Node 3 is start node and node 4 is final node. 1.62/0.82 || 1.62/0.82 || Those nodes are connected through the following edges: 1.62/0.82 || 1.62/0.82 || * 3 to 10 labelled f_1(0)* 3 to 19 labelled g_1(0)* 4 to 4 labelled #_1(0)* 10 to 11 labelled f_1(0)* 11 to 4 labelled g_1(0)* 11 to 20 labelled f_1(1)* 19 to 4 labelled h_1(0)* 19 to 22 labelled g_1(1)* 20 to 21 labelled f_1(1)* 21 to 4 labelled g_1(1)* 21 to 20 labelled f_1(1)* 22 to 4 labelled h_1(1)* 22 to 22 labelled g_1(1) 1.62/0.82 || 1.62/0.82 || 1.62/0.82 || ---------------------------------------- 1.62/0.82 || 1.62/0.82 || (4) 1.62/0.82 || YES 1.62/0.82 || 1.62/0.82 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]). 1.62/0.82 1.62/0.82 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 1.62/0.82 1.62/0.82 Dependency Pairs P_0: 1.62/0.82 1.62/0.82 1.62/0.82 Rules R_0: 1.62/0.82 1.62/0.82 f(g(X)) => g(f(f(X))) 1.62/0.82 f(h(X)) => h(g(X)) 1.62/0.82 1.62/0.82 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 1.62/0.82 1.62/0.82 We consider the dependency pair problem (P_0, R_0, static, formative). 1.62/0.82 1.62/0.82 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 1.62/0.82 1.62/0.82 1.62/0.82 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 1.62/0.82 1.62/0.82 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 1.62/0.82 1.62/0.82 1.62/0.82 +++ Citations +++ 1.62/0.82 1.62/0.82 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 1.62/0.82 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 1.62/0.82 [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. 1.62/0.82 EOF