/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 gcd : [o * o] --> o if : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o pred : [o] --> o s : [o] --> o true : [] --> o minus(X, s(Y)) => pred(minus(X, Y)) minus(X, 0) => X pred(s(X)) => X le(s(X), s(Y)) => le(X, Y) le(s(X), 0) => false le(0, X) => true gcd(0, X) => 0 gcd(s(X), 0) => s(X) gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if(false, s(X), s(Y)) => gcd(minus(Y, X), 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 : [] --> hd false : [] --> cc gcd : [hd * hd] --> hd if : [cc * hd * hd] --> hd le : [hd * hd] --> cc minus : [hd * hd] --> hd pred : [hd] --> hd s : [hd] --> hd true : [] --> cc 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] le#(s(X), s(Y)) =#> le#(X, Y) 3] gcd#(s(X), s(Y)) =#> if#(le(Y, X), s(X), s(Y)) 4] gcd#(s(X), s(Y)) =#> le#(Y, X) 5] if#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), s(Y)) 6] if#(true, s(X), s(Y)) =#> minus#(X, Y) 7] if#(false, s(X), s(Y)) =#> gcd#(minus(Y, X), s(X)) 8] if#(false, s(X), s(Y)) =#> minus#(Y, X) Rules R_0: minus(X, s(Y)) => pred(minus(X, Y)) minus(X, 0) => X pred(s(X)) => X le(s(X), s(Y)) => le(X, Y) le(s(X), 0) => false le(0, X) => true gcd(0, X) => 0 gcd(s(X), 0) => s(X) gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if(false, s(X), s(Y)) => gcd(minus(Y, X), 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 : * 1 : 0, 1 * 2 : 2 * 3 : 5, 6, 7, 8 * 4 : 2 * 5 : 3, 4 * 6 : 0, 1 * 7 : 3, 4 * 8 : 0, 1 This graph has the following strongly connected components: P_1: minus#(X, s(Y)) =#> minus#(X, Y) P_2: le#(s(X), s(Y)) =#> le#(X, Y) P_3: gcd#(s(X), s(Y)) =#> if#(le(Y, X), s(X), s(Y)) if#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), s(Y)) if#(false, s(X), s(Y)) =#> gcd#(minus(Y, X), s(X)) 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: minus(X, s(Y)) => pred(minus(X, Y)) minus(X, 0) => X pred(s(X)) => X le(s(X), s(Y)) => le(X, Y) le(s(X), 0) => false le(0, X) => true It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: gcd#(s(X), s(Y)) >? if#(le(Y, X), s(X), s(Y)) if#(true, s(X), s(Y)) >? gcd#(minus(X, Y), s(Y)) if#(false, s(X), s(Y)) >? gcd#(minus(Y, X), s(X)) minus(X, s(Y)) >= pred(minus(X, Y)) minus(X, 0) >= X pred(s(X)) >= X le(s(X), s(Y)) >= le(X, Y) le(s(X), 0) >= false le(0, X) >= true We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 false = 2 gcd# = \y0y1.y1 + 2y0 if# = \y0y1y2.y0 + y1 + y2 le = \y0y1.y1 minus = \y0y1.y0 pred = \y0.y0 s = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[gcd#(s(_x0), s(_x1))]] = 2x1 + 4x0 >= 2x1 + 3x0 = [[if#(le(_x1, _x0), s(_x0), s(_x1))]] [[if#(true, s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[gcd#(minus(_x0, _x1), s(_x1))]] [[if#(false, s(_x0), s(_x1))]] = 2 + 2x0 + 2x1 > 2x0 + 2x1 = [[gcd#(minus(_x1, _x0), s(_x0))]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[pred(s(_x0))]] = 2x0 >= x0 = [[_x0]] [[le(s(_x0), s(_x1))]] = 2x1 >= x1 = [[le(_x0, _x1)]] [[le(s(_x0), 0)]] = 3 >= 2 = [[false]] [[le(0, _x0)]] = x0 >= 0 = [[true]] 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: gcd#(s(X), s(Y)) =#> if#(le(Y, X), s(X), s(Y)) if#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_4, R_0) are: minus(X, s(Y)) => pred(minus(X, Y)) minus(X, 0) => X pred(s(X)) => X le(s(X), s(Y)) => le(X, Y) le(s(X), 0) => false le(0, X) => true It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: gcd#(s(X), s(Y)) >? if#(le(Y, X), s(X), s(Y)) if#(true, s(X), s(Y)) >? gcd#(minus(X, Y), s(Y)) minus(X, s(Y)) >= pred(minus(X, Y)) minus(X, 0) >= X pred(s(X)) >= X le(s(X), s(Y)) >= le(X, Y) le(s(X), 0) >= false le(0, X) >= true We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: if#(x_1,x_2,x_3) = if#(x_2x_3) This leaves the following ordering requirements: gcd#(s(X), s(Y)) >= if#(le(Y, X), s(X), s(Y)) if#(true, s(X), s(Y)) > gcd#(minus(X, Y), s(Y)) minus(X, s(Y)) >= pred(minus(X, Y)) minus(X, 0) >= X pred(s(X)) >= X The following interpretation satisfies the requirements: 0 = 3 false = 0 gcd# = \y0y1.y0 if# = \y0y1y2.y1 le = \y0y1.0 minus = \y0y1.y0 pred = \y0.y0 s = \y0.1 + y0 true = 0 Using this interpretation, the requirements translate to: [[gcd#(s(_x0), s(_x1))]] = 1 + x0 >= 1 + x0 = [[if#(le(_x1, _x0), s(_x0), s(_x1))]] [[if#(true, s(_x0), s(_x1))]] = 1 + x0 > x0 = [[gcd#(minus(_x0, _x1), s(_x1))]] [[minus(_x0, s(_x1))]] = x0 >= x0 = [[pred(minus(_x0, _x1))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[pred(s(_x0))]] = 1 + x0 >= x0 = [[_x0]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_0, minimal, formative) by (P_5, R_0, minimal, formative), where P_5 consists of: gcd#(s(X), s(Y)) =#> if#(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_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, 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(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_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(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 +++ [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.