/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o false : [] --> o if!6220mod : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o mod : [o * o] --> o s : [o] --> o true : [] --> o le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) mod(0, X) => 0 mod(s(X), 0) => 0 mod(s(X), s(Y)) => if!6220mod(le(Y, X), s(X), s(Y)) if!6220mod(true, s(X), s(Y)) => mod(minus(X, Y), s(Y)) if!6220mod(false, s(X), s(Y)) => s(X) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: 0 : [] --> vc false : [] --> ub if!6220mod : [ub * vc * vc] --> vc le : [vc * vc] --> ub minus : [vc * vc] --> vc mod : [vc * vc] --> vc s : [vc] --> vc true : [] --> ub 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: 0] le#(s(X), s(Y)) =#> le#(X, Y) 1] minus#(s(X), s(Y)) =#> minus#(X, Y) 2] mod#(s(X), s(Y)) =#> if!6220mod#(le(Y, X), s(X), s(Y)) 3] mod#(s(X), s(Y)) =#> le#(Y, X) 4] if!6220mod#(true, s(X), s(Y)) =#> mod#(minus(X, Y), s(Y)) 5] if!6220mod#(true, s(X), s(Y)) =#> minus#(X, Y) Rules R_0: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) mod(0, X) => 0 mod(s(X), 0) => 0 mod(s(X), s(Y)) => if!6220mod(le(Y, X), s(X), s(Y)) if!6220mod(true, s(X), s(Y)) => mod(minus(X, Y), s(Y)) if!6220mod(false, s(X), s(Y)) => 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: * 0 : 0 * 1 : 1 * 2 : 4, 5 * 3 : 0 * 4 : 2, 3 * 5 : 1 This graph has the following strongly connected components: P_1: le#(s(X), s(Y)) =#> le#(X, Y) P_2: minus#(s(X), s(Y)) =#> minus#(X, Y) P_3: mod#(s(X), s(Y)) =#> if!6220mod#(le(Y, X), s(X), s(Y)) if!6220mod#(true, s(X), s(Y)) =#> mod#(minus(X, Y), s(Y)) 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), (P_2, R_0, m, f) and (P_3, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_3, R_0) are: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: mod#(s(X), s(Y)) >? if!6220mod#(le(Y, X), s(X), s(Y)) if!6220mod#(true, s(X), s(Y)) >? mod#(minus(X, Y), s(Y)) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: if!6220mod#(x_1,x_2,x_3) = if!6220mod#(x_2x_3) This leaves the following ordering requirements: mod#(s(X), s(Y)) >= if!6220mod#(le(Y, X), s(X), s(Y)) if!6220mod#(true, s(X), s(Y)) > mod#(minus(X, Y), s(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) The following interpretation satisfies the requirements: 0 = 3 false = 0 if!6220mod# = \y0y1y2.y1 le = \y0y1.0 minus = \y0y1.y0 mod# = \y0y1.y0 s = \y0.2 + y0 true = 0 Using this interpretation, the requirements translate to: [[mod#(s(_x0), s(_x1))]] = 2 + x0 >= 2 + x0 = [[if!6220mod#(le(_x1, _x0), s(_x0), s(_x1))]] [[if!6220mod#(true, s(_x0), s(_x1))]] = 2 + x0 > x0 = [[mod#(minus(_x0, _x1), s(_x1))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 2 + x0 >= x0 = [[minus(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_4, R_0, minimal, formative), where P_4 consists of: mod#(s(X), s(Y)) =#> if!6220mod#(le(Y, X), s(X), s(Y)) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, 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: * 0 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(minus#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(le#) = 1 Thus, we can orient the dependency pairs as follows: nu(le#(s(X), s(Y))) = s(X) |> X = nu(le#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [FuhGieParSchSwi11] C. Fuhs, J. Giesl, M. Parting, P. Schneider-Kamp, and S. Swiderski. Proving Termination by Dependency Pairs and Inductive Theorem Proving. In volume 47(2) of Journal of Automated Reasoning. 133--160, 2011. [Gra95] B. Gramlich. Abstract Relations Between Restricted Termination and Confluence Properties of Rewrite Systems. In volume 24(1-2) of Fundamentae Informaticae. 3--23, 1995. [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.