/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: 0 : [] --> a bits : [a] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d half : [a] --> a map : [c -> c * d] --> d nil : [] --> d s : [a] --> a true : [] --> b Rules: half(0) => 0 half(s(0)) => 0 half(s(s(x))) => s(half(x)) bits(0) => 0 bits(s(x)) => s(bits(half(s(x)))) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, y) => filter(f, y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): half(0) >? 0 half(s(0)) >? 0 half(s(s(X))) >? s(half(X)) bits(0) >? 0 bits(s(X)) >? s(bits(half(s(X)))) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) filter(F, nil) >? nil filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) filter2(true, F, X, Y) >? cons(X, filter(F, Y)) filter2(false, F, X, Y) >? filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 bits = \y0.y0 cons = \y0y1.3 + y1 + 2y0 false = 3 filter = \G0y1.2y1 + G0(0) + 3y1G0(y1) filter2 = \y0G1y2y3.2y0 + 2y2 + 2y3 + 2G1(0) + 3y3G1(y3) half = \y0.y0 map = \G0y1.2 + 3y1 + G0(y1) + 2y1G0(y1) nil = 1 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[half(0)]] = 0 >= 0 = [[0]] [[half(s(0))]] = 0 >= 0 = [[0]] [[half(s(s(_x0)))]] = x0 >= x0 = [[s(half(_x0))]] [[bits(0)]] = 0 >= 0 = [[0]] [[bits(s(_x0))]] = x0 >= x0 = [[s(bits(half(s(_x0))))]] [[map(_F0, nil)]] = 5 + 3F0(1) > 1 = [[nil]] [[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))]] [[filter(_F0, nil)]] = 2 + F0(0) + 3F0(1) > 1 = [[nil]] [[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)]] [[filter2(true, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 3 + 2x1 + 2x2 + F0(0) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 2x2 + F0(0) + 3x2F0(x2) = [[filter(_F0, _x2)]] We can thus remove the following rules: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): half(0) >? 0 half(s(0)) >? 0 half(s(s(X))) >? s(half(X)) bits(0) >? 0 bits(s(X)) >? s(bits(half(s(X)))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 bits = \y0.1 + y0 half = \y0.y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[half(0)]] = 0 >= 0 = [[0]] [[half(s(0))]] = 0 >= 0 = [[0]] [[half(s(s(_x0)))]] = x0 >= x0 = [[s(half(_x0))]] [[bits(0)]] = 1 > 0 = [[0]] [[bits(s(_x0))]] = 1 + x0 >= 1 + x0 = [[s(bits(half(s(_x0))))]] We can thus remove the following rules: bits(0) => 0 We observe that the rules contain a first-order subset: half(0) => 0 half(s(0)) => 0 half(s(s(X))) => s(half(X)) bits(s(X)) => s(bits(half(s(X)))) 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) QTRS Reverse [EQUIVALENT] || (2) QTRS || (3) RFCMatchBoundsTRSProof [EQUIVALENT] || (4) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || half(0) -> 0 || half(s(0)) -> 0 || half(s(s(%X))) -> s(half(%X)) || bits(s(%X)) -> s(bits(half(s(%X)))) || || Q is empty. || || ---------------------------------------- || || (1) QTRS Reverse (EQUIVALENT) || We applied the QTRS Reverse Processor [REVERSE]. || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || 0'(half(x)) -> 0'(x) || 0'(s(half(x))) -> 0'(x) || s(s(half(x))) -> half(s(x)) || s(bits(x)) -> s(half(bits(s(x)))) || || Q is empty. || || ---------------------------------------- || || (3) RFCMatchBoundsTRSProof (EQUIVALENT) || Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R. || The following rules were used to construct the certificate: || || 0'(half(x)) -> 0'(x) || 0'(s(half(x))) -> 0'(x) || s(s(half(x))) -> half(s(x)) || s(bits(x)) -> s(half(bits(s(x)))) || || The certificate found is represented by the following graph. || The certificate consists of the following enumerated nodes: || 2, 8, 9, 10, 11, 12, 13, 14, 15, 16 || || Node 2 is start node and node 8 is final node. || || Those nodes are connected through the following edges: || || * 2 to 8 labelled 0'_1(0), 0'_1(1)* 2 to 9 labelled half_1(0)* 2 to 10 labelled s_1(0)* 8 to 8 labelled #_1(0)* 9 to 8 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 8 labelled s_1(0)* 12 to 13 labelled half_1(1)* 12 to 14 labelled s_1(1)* 13 to 8 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 8 labelled s_1(1)* 16 to 13 labelled half_1(1)* 16 to 14 labelled s_1(1) || || || ---------------------------------------- || || (4) || 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 in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: Rules R_0: half(0) => 0 half(s(0)) => 0 half(s(s(X))) => s(half(X)) bits(s(X)) => s(bits(half(s(X)))) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 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 +++ [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.