1.69/0.89 YES 1.69/0.90 We consider the system theBenchmark. 1.69/0.90 1.69/0.90 Alphabet: 1.69/0.90 1.69/0.90 0 : [] --> a 1.69/0.90 bits : [a] --> a 1.69/0.90 cons : [c * d] --> d 1.69/0.90 false : [] --> b 1.69/0.90 filter : [c -> b * d] --> d 1.69/0.90 filter2 : [b * c -> b * c * d] --> d 1.69/0.90 half : [a] --> a 1.69/0.90 map : [c -> c * d] --> d 1.69/0.90 nil : [] --> d 1.69/0.90 s : [a] --> a 1.69/0.90 true : [] --> b 1.69/0.90 1.69/0.90 Rules: 1.69/0.90 1.69/0.90 half(0) => 0 1.69/0.90 half(s(0)) => 0 1.69/0.90 half(s(s(x))) => s(half(x)) 1.69/0.90 bits(0) => 0 1.69/0.90 bits(s(x)) => s(bits(half(s(x)))) 1.69/0.90 map(f, nil) => nil 1.69/0.90 map(f, cons(x, y)) => cons(f x, map(f, y)) 1.69/0.90 filter(f, nil) => nil 1.69/0.90 filter(f, cons(x, y)) => filter2(f x, f, x, y) 1.69/0.90 filter2(true, f, x, y) => cons(x, filter(f, y)) 1.69/0.90 filter2(false, f, x, y) => filter(f, y) 1.69/0.90 1.69/0.90 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 1.69/0.90 1.69/0.90 We use rule removal, following [Kop12, Theorem 2.23]. 1.69/0.90 1.69/0.90 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 1.69/0.90 1.69/0.90 half(0) >? 0 1.69/0.90 half(s(0)) >? 0 1.69/0.90 half(s(s(X))) >? s(half(X)) 1.69/0.90 bits(0) >? 0 1.69/0.90 bits(s(X)) >? s(bits(half(s(X)))) 1.69/0.90 map(F, nil) >? nil 1.69/0.90 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 1.69/0.90 filter(F, nil) >? nil 1.69/0.90 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 1.69/0.90 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 1.69/0.90 filter2(false, F, X, Y) >? filter(F, Y) 1.69/0.90 1.69/0.90 We orient these requirements with a polynomial interpretation in the natural numbers. 1.69/0.90 1.69/0.90 The following interpretation satisfies the requirements: 1.69/0.90 1.69/0.90 0 = 0 1.69/0.90 bits = \y0.y0 1.69/0.90 cons = \y0y1.3 + y1 + 2y0 1.69/0.90 false = 3 1.69/0.90 filter = \G0y1.2y1 + G0(0) + 3y1G0(y1) 1.69/0.90 filter2 = \y0G1y2y3.2y0 + 2y2 + 2y3 + 2G1(0) + 3y3G1(y3) 1.69/0.90 half = \y0.y0 1.69/0.90 map = \G0y1.2 + 3y1 + G0(y1) + 2y1G0(y1) 1.69/0.90 nil = 1 1.69/0.90 s = \y0.y0 1.69/0.90 true = 3 1.69/0.90 1.69/0.90 Using this interpretation, the requirements translate to: 1.69/0.90 1.69/0.90 [[half(0)]] = 0 >= 0 = [[0]] 1.69/0.90 [[half(s(0))]] = 0 >= 0 = [[0]] 1.69/0.90 [[half(s(s(_x0)))]] = x0 >= x0 = [[s(half(_x0))]] 1.69/0.90 [[bits(0)]] = 0 >= 0 = [[0]] 1.69/0.90 [[bits(s(_x0))]] = x0 >= x0 = [[s(bits(half(s(_x0))))]] 1.69/0.90 [[map(_F0, nil)]] = 5 + 3F0(1) > 1 = [[nil]] 1.69/0.90 [[map(_F0, cons(_x1, _x2))]] = 11 + 3x2 + 6x1 + 2x2F0(3 + x2 + 2x1) + 4x1F0(3 + x2 + 2x1) + 7F0(3 + x2 + 2x1) > 5 + 2x1 + 3x2 + F0(x2) + 2x2F0(x2) + 2F0(x1) = [[cons(_F0 _x1, map(_F0, _x2))]] 1.69/0.90 [[filter(_F0, nil)]] = 2 + F0(0) + 3F0(1) > 1 = [[nil]] 1.69/0.90 [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + F0(0) + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) > 2x2 + 4x1 + 2F0(0) + 2F0(x1) + 3x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 1.69/0.90 [[filter2(true, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 3 + 2x1 + 2x2 + F0(0) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 1.69/0.90 [[filter2(false, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 2x2 + F0(0) + 3x2F0(x2) = [[filter(_F0, _x2)]] 1.69/0.90 1.69/0.90 We can thus remove the following rules: 1.69/0.90 1.69/0.90 map(F, nil) => nil 1.69/0.90 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 1.69/0.90 filter(F, nil) => nil 1.69/0.90 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 1.69/0.90 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 1.69/0.90 filter2(false, F, X, Y) => filter(F, Y) 1.69/0.90 1.69/0.90 We use rule removal, following [Kop12, Theorem 2.23]. 1.69/0.90 1.69/0.90 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 1.69/0.90 1.69/0.90 half(0) >? 0 1.69/0.90 half(s(0)) >? 0 1.69/0.90 half(s(s(X))) >? s(half(X)) 1.69/0.90 bits(0) >? 0 1.69/0.90 bits(s(X)) >? s(bits(half(s(X)))) 1.69/0.90 1.69/0.90 We orient these requirements with a polynomial interpretation in the natural numbers. 1.69/0.90 1.69/0.90 The following interpretation satisfies the requirements: 1.69/0.90 1.69/0.90 0 = 0 1.69/0.90 bits = \y0.1 + y0 1.69/0.90 half = \y0.y0 1.69/0.90 s = \y0.y0 1.69/0.90 1.69/0.90 Using this interpretation, the requirements translate to: 1.69/0.90 1.69/0.90 [[half(0)]] = 0 >= 0 = [[0]] 1.69/0.90 [[half(s(0))]] = 0 >= 0 = [[0]] 1.69/0.90 [[half(s(s(_x0)))]] = x0 >= x0 = [[s(half(_x0))]] 1.69/0.90 [[bits(0)]] = 1 > 0 = [[0]] 1.69/0.90 [[bits(s(_x0))]] = 1 + x0 >= 1 + x0 = [[s(bits(half(s(_x0))))]] 1.69/0.90 1.69/0.90 We can thus remove the following rules: 1.69/0.90 1.69/0.90 bits(0) => 0 1.69/0.90 1.69/0.90 We observe that the rules contain a first-order subset: 1.69/0.90 1.69/0.90 half(0) => 0 1.69/0.90 half(s(0)) => 0 1.69/0.90 half(s(s(X))) => s(half(X)) 1.69/0.90 bits(s(X)) => s(bits(half(s(X)))) 1.69/0.90 1.69/0.90 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. 1.69/0.90 1.69/0.90 According to the external first-order termination prover, this system is indeed terminating: 1.69/0.90 1.69/0.90 || proof of resources/system.trs 1.69/0.90 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.69/0.90 || 1.69/0.90 || 1.69/0.90 || Termination w.r.t. Q of the given QTRS could be proven: 1.69/0.90 || 1.69/0.90 || (0) QTRS 1.69/0.90 || (1) QTRS Reverse [EQUIVALENT] 1.69/0.90 || (2) QTRS 1.69/0.90 || (3) RFCMatchBoundsTRSProof [EQUIVALENT] 1.69/0.90 || (4) YES 1.69/0.90 || 1.69/0.90 || 1.69/0.90 || ---------------------------------------- 1.69/0.90 || 1.69/0.90 || (0) 1.69/0.90 || Obligation: 1.69/0.90 || Q restricted rewrite system: 1.69/0.90 || The TRS R consists of the following rules: 1.69/0.90 || 1.69/0.90 || half(0) -> 0 1.69/0.90 || half(s(0)) -> 0 1.69/0.90 || half(s(s(%X))) -> s(half(%X)) 1.69/0.90 || bits(s(%X)) -> s(bits(half(s(%X)))) 1.69/0.90 || 1.69/0.90 || Q is empty. 1.69/0.90 || 1.69/0.90 || ---------------------------------------- 1.69/0.90 || 1.69/0.90 || (1) QTRS Reverse (EQUIVALENT) 1.69/0.90 || We applied the QTRS Reverse Processor [REVERSE]. 1.69/0.90 || ---------------------------------------- 1.69/0.90 || 1.69/0.90 || (2) 1.69/0.90 || Obligation: 1.69/0.90 || Q restricted rewrite system: 1.69/0.90 || The TRS R consists of the following rules: 1.69/0.90 || 1.69/0.90 || 0'(half(x)) -> 0'(x) 1.69/0.90 || 0'(s(half(x))) -> 0'(x) 1.69/0.90 || s(s(half(x))) -> half(s(x)) 1.69/0.90 || s(bits(x)) -> s(half(bits(s(x)))) 1.69/0.90 || 1.69/0.90 || Q is empty. 1.69/0.90 || 1.69/0.90 || ---------------------------------------- 1.69/0.90 || 1.69/0.90 || (3) RFCMatchBoundsTRSProof (EQUIVALENT) 1.69/0.90 || Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R. 1.69/0.90 || The following rules were used to construct the certificate: 1.69/0.90 || 1.69/0.90 || 0'(half(x)) -> 0'(x) 1.69/0.90 || 0'(s(half(x))) -> 0'(x) 1.69/0.90 || s(s(half(x))) -> half(s(x)) 1.69/0.90 || s(bits(x)) -> s(half(bits(s(x)))) 1.69/0.90 || 1.69/0.90 || The certificate found is represented by the following graph. 1.69/0.90 || The certificate consists of the following enumerated nodes: 1.69/0.90 || 3, 4, 9, 10, 11, 12, 13, 14, 15, 16 1.69/0.90 || 1.69/0.90 || Node 3 is start node and node 4 is final node. 1.69/0.90 || 1.69/0.90 || Those nodes are connected through the following edges: 1.69/0.90 || 1.69/0.90 || * 3 to 4 labelled 0'_1(0), 0'_1(1)* 3 to 9 labelled half_1(0)* 3 to 10 labelled s_1(0)* 4 to 4 labelled #_1(0)* 9 to 4 labelled s_1(0)* 9 to 13 labelled half_1(1)* 9 to 14 labelled s_1(1)* 10 to 11 labelled half_1(0)* 11 to 12 labelled bits_1(0)* 12 to 4 labelled s_1(0)* 12 to 13 labelled half_1(1)* 12 to 14 labelled s_1(1)* 13 to 4 labelled s_1(1)* 13 to 13 labelled half_1(1)* 13 to 14 labelled s_1(1)* 14 to 15 labelled half_1(1)* 15 to 16 labelled bits_1(1)* 16 to 4 labelled s_1(1)* 16 to 13 labelled half_1(1)* 16 to 14 labelled s_1(1) 1.69/0.90 || 1.69/0.90 || 1.69/0.90 || ---------------------------------------- 1.69/0.90 || 1.69/0.90 || (4) 1.69/0.90 || YES 1.69/0.90 || 1.69/0.90 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]). 1.69/0.90 1.69/0.90 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 1.69/0.90 1.69/0.90 Dependency Pairs P_0: 1.69/0.90 1.69/0.90 1.69/0.90 Rules R_0: 1.69/0.90 1.69/0.90 half(0) => 0 1.69/0.90 half(s(0)) => 0 1.69/0.90 half(s(s(X))) => s(half(X)) 1.69/0.90 bits(s(X)) => s(bits(half(s(X)))) 1.69/0.90 1.69/0.90 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 1.69/0.90 1.69/0.90 We consider the dependency pair problem (P_0, R_0, static, formative). 1.69/0.90 1.69/0.90 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 1.69/0.90 1.69/0.90 1.69/0.90 This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 1.69/0.90 1.69/0.90 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 1.69/0.90 1.69/0.90 1.69/0.90 +++ Citations +++ 1.69/0.90 1.69/0.90 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 1.69/0.90 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 1.69/0.90 [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. 1.69/0.90 EOF