/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o minus : [o * o] --> o pred : [o] --> o quot : [o * o] --> o s : [o] --> o pred(s(X)) => X minus(X, 0) => X minus(X, s(Y)) => pred(minus(X, Y)) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 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] minus#(X, s(Y)) =#> pred#(minus(X, Y)) 1] minus#(X, s(Y)) =#> minus#(X, Y) 2] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 3] quot#(s(X), s(Y)) =#> minus#(X, Y) Rules R_0: pred(s(X)) => X minus(X, 0) => X minus(X, s(Y)) => pred(minus(X, Y)) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 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 : * 1 : 0, 1 * 2 : 2, 3 * 3 : 0, 1 This graph has the following strongly connected components: P_1: minus#(X, s(Y)) =#> minus#(X, Y) P_2: quot#(s(X), s(Y)) =#> quot#(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) and (P_2, R_0, m, f). 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_2, R_0) are: pred(s(X)) => X minus(X, 0) => X minus(X, s(Y)) => pred(minus(X, Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: quot#(s(X), s(Y)) >? quot#(minus(X, Y), s(Y)) pred(s(X)) >= X minus(X, 0) >= X minus(X, s(Y)) >= pred(minus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 minus = \y0y1.1 + y0 pred = \y0.y0 quot# = \y0y1.y0 s = \y0.3 + y0 Using this interpretation, the requirements translate to: [[quot#(s(_x0), s(_x1))]] = 3 + x0 > 1 + x0 = [[quot#(minus(_x0, _x1), s(_x1))]] [[pred(s(_x0))]] = 3 + x0 >= x0 = [[_x0]] [[minus(_x0, 0)]] = 1 + x0 >= x0 = [[_x0]] [[minus(_x0, s(_x1))]] = 1 + x0 >= 1 + x0 = [[pred(minus(_x0, _x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0). 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(minus#) = 2 Thus, we can orient the dependency pairs as follows: nu(minus#(X, s(Y))) = s(Y) |> Y = nu(minus#(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 +++ [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.