1.33/0.77 YES 1.33/0.78 We consider the system theBenchmark. 1.33/0.78 1.33/0.78 Alphabet: 1.33/0.78 1.33/0.78 a : [] --> a 1.33/0.78 b : [] --> a 1.33/0.78 cons : [e * f] --> f 1.33/0.78 f : [a] --> b 1.33/0.78 false : [] --> d 1.33/0.78 filter : [e -> d * f] --> f 1.33/0.78 filter2 : [d * e -> d * e * f] --> f 1.33/0.78 g : [a] --> c 1.33/0.78 map : [e -> e * f] --> f 1.33/0.78 nil : [] --> f 1.33/0.78 true : [] --> d 1.33/0.78 1.33/0.78 Rules: 1.33/0.78 1.33/0.78 f(a) => f(b) 1.33/0.78 g(b) => g(a) 1.33/0.78 map(h, nil) => nil 1.33/0.78 map(h, cons(x, y)) => cons(h x, map(h, y)) 1.33/0.78 filter(h, nil) => nil 1.33/0.78 filter(h, cons(x, y)) => filter2(h x, h, x, y) 1.33/0.78 filter2(true, h, x, y) => cons(x, filter(h, y)) 1.33/0.78 filter2(false, h, x, y) => filter(h, y) 1.33/0.78 1.33/0.78 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 1.33/0.78 1.33/0.78 We use rule removal, following [Kop12, Theorem 2.23]. 1.33/0.78 1.33/0.78 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 1.33/0.78 1.33/0.78 f(a) >? f(b) 1.33/0.78 g(b) >? g(a) 1.33/0.78 map(F, nil) >? nil 1.33/0.78 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 1.33/0.78 filter(F, nil) >? nil 1.33/0.78 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 1.33/0.78 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 1.33/0.78 filter2(false, F, X, Y) >? filter(F, Y) 1.33/0.78 1.33/0.78 We orient these requirements with a polynomial interpretation in the natural numbers. 1.33/0.78 1.33/0.78 The following interpretation satisfies the requirements: 1.33/0.78 1.33/0.78 a = 0 1.33/0.78 b = 0 1.33/0.78 cons = \y0y1.2 + y1 + 2y0 1.33/0.78 f = \y0.y0 1.33/0.78 false = 3 1.33/0.78 filter = \G0y1.2 + 2y1 + G0(0) + y1G0(y1) 1.33/0.78 filter2 = \y0G1y2y3.1 + y0 + 2y3 + 3y2 + G1(y2) + y3G1(y3) 1.33/0.78 g = \y0.2y0 1.33/0.78 map = \G0y1.1 + 3y1 + G0(0) + 2y1G0(y1) 1.33/0.78 nil = 0 1.33/0.78 true = 3 1.33/0.78 1.33/0.78 Using this interpretation, the requirements translate to: 1.33/0.78 1.33/0.78 [[f(a)]] = 0 >= 0 = [[f(b)]] 1.33/0.78 [[g(b)]] = 0 >= 0 = [[g(a)]] 1.33/0.78 [[map(_F0, nil)]] = 1 + F0(0) > 0 = [[nil]] 1.33/0.78 [[map(_F0, cons(_x1, _x2))]] = 7 + 3x2 + 6x1 + F0(0) + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) > 3 + 2x1 + 3x2 + F0(0) + 2x2F0(x2) + 2F0(x1) = [[cons(_F0 _x1, map(_F0, _x2))]] 1.33/0.78 [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 1.33/0.78 [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + F0(0) + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > 1 + 2x2 + 4x1 + 2F0(x1) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 1.33/0.78 [[filter2(true, _F0, _x1, _x2)]] = 4 + 2x2 + 3x1 + F0(x1) + x2F0(x2) >= 4 + 2x1 + 2x2 + F0(0) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 1.33/0.78 [[filter2(false, _F0, _x1, _x2)]] = 4 + 2x2 + 3x1 + F0(x1) + x2F0(x2) > 2 + 2x2 + F0(0) + x2F0(x2) = [[filter(_F0, _x2)]] 1.33/0.78 1.33/0.78 We can thus remove the following rules: 1.33/0.78 1.33/0.78 map(F, nil) => nil 1.33/0.78 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 1.33/0.78 filter(F, nil) => nil 1.33/0.78 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 1.33/0.78 filter2(false, F, X, Y) => filter(F, Y) 1.33/0.78 1.33/0.78 We use rule removal, following [Kop12, Theorem 2.23]. 1.33/0.78 1.33/0.78 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 1.33/0.78 1.33/0.78 f(a) >? f(b) 1.33/0.78 g(b) >? g(a) 1.33/0.78 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 1.33/0.78 1.33/0.78 We orient these requirements with a polynomial interpretation in the natural numbers. 1.33/0.78 1.33/0.78 The following interpretation satisfies the requirements: 1.33/0.78 1.33/0.78 a = 0 1.33/0.78 b = 0 1.33/0.78 cons = \y0y1.y0 + y1 1.33/0.78 f = \y0.2y0 1.33/0.78 filter = \G0y1.y1 + G0(0) 1.33/0.78 filter2 = \y0G1y2y3.3 + 3y0 + 3y2 + 3y3 + G1(y0) + 2G1(0) + 2G1(y3) 1.33/0.78 g = \y0.y0 1.33/0.78 true = 3 1.33/0.78 1.33/0.78 Using this interpretation, the requirements translate to: 1.33/0.78 1.33/0.78 [[f(a)]] = 0 >= 0 = [[f(b)]] 1.33/0.78 [[g(b)]] = 0 >= 0 = [[g(a)]] 1.33/0.78 [[filter2(true, _F0, _x1, _x2)]] = 12 + 3x1 + 3x2 + F0(3) + 2F0(0) + 2F0(x2) > x1 + x2 + F0(0) = [[cons(_x1, filter(_F0, _x2))]] 1.33/0.78 1.33/0.78 We can thus remove the following rules: 1.33/0.78 1.33/0.78 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 1.33/0.78 1.33/0.78 We observe that the rules contain a first-order subset: 1.33/0.78 1.33/0.78 f(a) => f(b) 1.33/0.78 g(b) => g(a) 1.33/0.78 1.33/0.78 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. 1.33/0.78 1.33/0.78 According to the external first-order termination prover, this system is indeed terminating: 1.33/0.78 1.33/0.78 || proof of resources/system.trs 1.33/0.78 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.33/0.78 || 1.33/0.78 || 1.33/0.78 || Termination w.r.t. Q of the given QTRS could be proven: 1.33/0.78 || 1.33/0.78 || (0) QTRS 1.33/0.78 || (1) QTRS Reverse [EQUIVALENT] 1.33/0.78 || (2) QTRS 1.33/0.78 || (3) RFCMatchBoundsTRSProof [EQUIVALENT] 1.33/0.78 || (4) YES 1.33/0.78 || 1.33/0.78 || 1.33/0.78 || ---------------------------------------- 1.33/0.78 || 1.33/0.78 || (0) 1.33/0.78 || Obligation: 1.33/0.78 || Q restricted rewrite system: 1.33/0.78 || The TRS R consists of the following rules: 1.33/0.78 || 1.33/0.78 || f(a) -> f(b) 1.33/0.78 || g(b) -> g(a) 1.33/0.78 || 1.33/0.78 || Q is empty. 1.33/0.78 || 1.33/0.78 || ---------------------------------------- 1.33/0.78 || 1.33/0.78 || (1) QTRS Reverse (EQUIVALENT) 1.33/0.78 || We applied the QTRS Reverse Processor [REVERSE]. 1.33/0.78 || ---------------------------------------- 1.33/0.78 || 1.33/0.78 || (2) 1.33/0.78 || Obligation: 1.33/0.78 || Q restricted rewrite system: 1.33/0.78 || The TRS R consists of the following rules: 1.33/0.78 || 1.33/0.78 || a'(f(x)) -> b'(f(x)) 1.33/0.78 || b'(g(x)) -> a'(g(x)) 1.33/0.78 || 1.33/0.78 || Q is empty. 1.33/0.78 || 1.33/0.78 || ---------------------------------------- 1.33/0.78 || 1.33/0.78 || (3) RFCMatchBoundsTRSProof (EQUIVALENT) 1.33/0.78 || Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 0. This implies Q-termination of R. 1.33/0.78 || The following rules were used to construct the certificate: 1.33/0.78 || 1.33/0.78 || a'(f(x)) -> b'(f(x)) 1.33/0.78 || b'(g(x)) -> a'(g(x)) 1.33/0.78 || 1.33/0.78 || The certificate found is represented by the following graph. 1.33/0.78 || The certificate consists of the following enumerated nodes: 1.33/0.78 || 2, 4, 5, 6 1.33/0.78 || 1.33/0.78 || Node 2 is start node and node 4 is final node. 1.33/0.78 || 1.33/0.78 || Those nodes are connected through the following edges: 1.33/0.78 || 1.33/0.78 || * 2 to 5 labelled b'_1(0)* 2 to 6 labelled a'_1(0)* 4 to 4 labelled #_1(0)* 5 to 4 labelled f_1(0)* 6 to 4 labelled g_1(0) 1.33/0.78 || 1.33/0.78 || 1.33/0.78 || ---------------------------------------- 1.33/0.78 || 1.33/0.78 || (4) 1.33/0.78 || YES 1.33/0.78 || 1.33/0.78 We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). 1.33/0.78 1.33/0.78 We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): 1.33/0.78 1.33/0.78 Dependency Pairs P_0: 1.33/0.78 1.33/0.78 1.33/0.78 Rules R_0: 1.33/0.78 1.33/0.78 f(a) => f(b) 1.33/0.78 g(b) => g(a) 1.33/0.78 1.33/0.78 Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. 1.33/0.78 1.33/0.78 We consider the dependency pair problem (P_0, R_0, minimal, formative). 1.33/0.78 1.33/0.78 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 1.33/0.78 1.33/0.78 1.33/0.78 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 1.33/0.78 1.33/0.78 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 1.33/0.78 1.33/0.78 1.33/0.78 +++ Citations +++ 1.33/0.78 1.33/0.78 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 1.33/0.78 [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. 1.33/0.78 EOF