/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. F : [] --> o G : [] --> o app : [o * o] --> o app(app(F, app(app(F, X), Y)), Y) => app(app(F, app(G, app(app(F, X), Y))), app(X, 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, all): Dependency Pairs P_0: 0] app#(app(F, app(app(F, X), Y)), Y) =#> app#(app(F, app(G, app(app(F, X), Y))), app(X, Y)) 1] app#(app(F, app(app(F, X), Y)), Y) =#> app#(F, app(G, app(app(F, X), Y))) 2] app#(app(F, app(app(F, X), Y)), Y) =#> app#(G, app(app(F, X), Y)) 3] app#(app(F, app(app(F, X), Y)), Y) =#> app#(app(F, X), Y) 4] app#(app(F, app(app(F, X), Y)), Y) =#> app#(F, X) 5] app#(app(F, app(app(F, X), Y)), Y) =#> app#(X, Y) Rules R_0: app(app(F, app(app(F, X), Y)), Y) => app(app(F, app(G, app(app(F, X), Y))), app(X, Y)) Thus, the original system is terminating if (P_0, R_0, minimal, all) is finite. We consider the dependency pair problem (P_0, R_0, minimal, all). 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, 2, 3, 4, 5 * 1 : * 2 : * 3 : 0, 1, 2, 3, 4, 5 * 4 : * 5 : 0, 1, 2, 3, 4, 5 This graph has the following strongly connected components: P_1: app#(app(F, app(app(F, X), Y)), Y) =#> app#(app(F, app(G, app(app(F, X), Y))), app(X, Y)) app#(app(F, app(app(F, X), Y)), Y) =#> app#(app(F, X), Y) app#(app(F, app(app(F, X), Y)), Y) =#> app#(X, 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). Thus, the original system is terminating if (P_1, R_0, minimal, all) is finite. We consider the dependency pair problem (P_1, R_0, minimal, all). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: app#(app(F, app(app(F, X), Y)), Y) >? app#(app(F, app(G, app(app(F, X), Y))), app(X, Y)) app#(app(F, app(app(F, X), Y)), Y) >? app#(app(F, X), Y) app#(app(F, app(app(F, X), Y)), Y) >? app#(X, Y) app(app(F, app(app(F, X), Y)), Y) >= app(app(F, app(G, app(app(F, X), Y))), app(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: F = 2 G = 0 app = \y0y1.y0 + y0y1 app# = \y0y1.2y0 Using this interpretation, the requirements translate to: [[app#(app(F, app(app(F, _x0), _x1)), _x1)]] = 12 + 8x0 + 8x0x1 + 8x1 > 4 = [[app#(app(F, app(G, app(app(F, _x0), _x1))), app(_x0, _x1))]] [[app#(app(F, app(app(F, _x0), _x1)), _x1)]] = 12 + 8x0 + 8x0x1 + 8x1 > 4 + 4x0 = [[app#(app(F, _x0), _x1)]] [[app#(app(F, app(app(F, _x0), _x1)), _x1)]] = 12 + 8x0 + 8x0x1 + 8x1 > 2x0 = [[app#(_x0, _x1)]] [[app(app(F, app(app(F, _x0), _x1)), _x1)]] = 6 + 4x0 + 4x0x1x1 + 4x1x1 + 8x0x1 + 10x1 >= 2 + 2x0 + 2x0x1 = [[app(app(F, app(G, app(app(F, _x0), _x1))), app(_x0, _x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0). 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.