/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 cons : [o * o] --> o eq : [o * o] --> o false : [] --> o ifmin : [o * o] --> o ifrepl : [o * o * o * o] --> o ifselsort : [o * o] --> o le : [o * o] --> o min : [o] --> o nil : [] --> o replace : [o * o * o] --> o s : [o] --> o selsort : [o] --> o true : [] --> o eq(0, 0) => true eq(0, s(X)) => false eq(s(X), 0) => false eq(s(X), s(Y)) => eq(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) min(cons(0, nil)) => 0 min(cons(s(X), nil)) => s(X) min(cons(X, cons(Y, Z))) => ifmin(le(X, Y), cons(X, cons(Y, Z))) ifmin(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) ifmin(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) replace(X, Y, nil) => nil replace(X, Y, cons(Z, U)) => ifrepl(eq(X, Z), X, Y, cons(Z, U)) ifrepl(true, X, Y, cons(Z, U)) => cons(Y, U) ifrepl(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) selsort(nil) => nil selsort(cons(X, Y)) => ifselsort(eq(X, min(cons(X, Y))), cons(X, Y)) ifselsort(true, cons(X, Y)) => cons(X, selsort(Y)) ifselsort(false, cons(X, Y)) => cons(min(cons(X, Y)), selsort(replace(min(cons(X, Y)), X, Y))) 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 : [] --> uh cons : [uh * zh] --> zh eq : [uh * uh] --> sg false : [] --> sg ifmin : [sg * zh] --> uh ifrepl : [sg * uh * uh * zh] --> zh ifselsort : [sg * zh] --> zh le : [uh * uh] --> sg min : [zh] --> uh nil : [] --> zh replace : [uh * uh * zh] --> zh s : [uh] --> uh selsort : [zh] --> zh true : [] --> sg 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] eq#(s(X), s(Y)) =#> eq#(X, Y) 1] le#(s(X), s(Y)) =#> le#(X, Y) 2] min#(cons(X, cons(Y, Z))) =#> ifmin#(le(X, Y), cons(X, cons(Y, Z))) 3] min#(cons(X, cons(Y, Z))) =#> le#(X, Y) 4] ifmin#(true, cons(X, cons(Y, Z))) =#> min#(cons(X, Z)) 5] ifmin#(false, cons(X, cons(Y, Z))) =#> min#(cons(Y, Z)) 6] replace#(X, Y, cons(Z, U)) =#> ifrepl#(eq(X, Z), X, Y, cons(Z, U)) 7] replace#(X, Y, cons(Z, U)) =#> eq#(X, Z) 8] ifrepl#(false, X, Y, cons(Z, U)) =#> replace#(X, Y, U) 9] selsort#(cons(X, Y)) =#> ifselsort#(eq(X, min(cons(X, Y))), cons(X, Y)) 10] selsort#(cons(X, Y)) =#> eq#(X, min(cons(X, Y))) 11] selsort#(cons(X, Y)) =#> min#(cons(X, Y)) 12] ifselsort#(true, cons(X, Y)) =#> selsort#(Y) 13] ifselsort#(false, cons(X, Y)) =#> min#(cons(X, Y)) 14] ifselsort#(false, cons(X, Y)) =#> selsort#(replace(min(cons(X, Y)), X, Y)) 15] ifselsort#(false, cons(X, Y)) =#> replace#(min(cons(X, Y)), X, Y) 16] ifselsort#(false, cons(X, Y)) =#> min#(cons(X, Y)) Rules R_0: eq(0, 0) => true eq(0, s(X)) => false eq(s(X), 0) => false eq(s(X), s(Y)) => eq(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) min(cons(0, nil)) => 0 min(cons(s(X), nil)) => s(X) min(cons(X, cons(Y, Z))) => ifmin(le(X, Y), cons(X, cons(Y, Z))) ifmin(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) ifmin(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) replace(X, Y, nil) => nil replace(X, Y, cons(Z, U)) => ifrepl(eq(X, Z), X, Y, cons(Z, U)) ifrepl(true, X, Y, cons(Z, U)) => cons(Y, U) ifrepl(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) selsort(nil) => nil selsort(cons(X, Y)) => ifselsort(eq(X, min(cons(X, Y))), cons(X, Y)) ifselsort(true, cons(X, Y)) => cons(X, selsort(Y)) ifselsort(false, cons(X, Y)) => cons(min(cons(X, Y)), selsort(replace(min(cons(X, Y)), X, 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 : 0 * 1 : 1 * 2 : 4, 5 * 3 : 1 * 4 : 2, 3 * 5 : 2, 3 * 6 : 8 * 7 : 0 * 8 : 6, 7 * 9 : 12, 13, 14, 15, 16 * 10 : 0 * 11 : 2, 3 * 12 : 9, 10, 11 * 13 : 2, 3 * 14 : 9, 10, 11 * 15 : 6, 7 * 16 : 2, 3 This graph has the following strongly connected components: P_1: eq#(s(X), s(Y)) =#> eq#(X, Y) P_2: le#(s(X), s(Y)) =#> le#(X, Y) P_3: min#(cons(X, cons(Y, Z))) =#> ifmin#(le(X, Y), cons(X, cons(Y, Z))) ifmin#(true, cons(X, cons(Y, Z))) =#> min#(cons(X, Z)) ifmin#(false, cons(X, cons(Y, Z))) =#> min#(cons(Y, Z)) P_4: replace#(X, Y, cons(Z, U)) =#> ifrepl#(eq(X, Z), X, Y, cons(Z, U)) ifrepl#(false, X, Y, cons(Z, U)) =#> replace#(X, Y, U) P_5: selsort#(cons(X, Y)) =#> ifselsort#(eq(X, min(cons(X, Y))), cons(X, Y)) ifselsort#(true, cons(X, Y)) =#> selsort#(Y) ifselsort#(false, cons(X, Y)) =#> selsort#(replace(min(cons(X, Y)), 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), (P_2, R_0, m, f), (P_3, R_0, m, f), (P_4, R_0, m, f) and (P_5, 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), (P_3, R_0, minimal, formative), (P_4, 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_5, R_0) are: eq(0, 0) => true eq(0, s(X)) => false eq(s(X), 0) => false eq(s(X), s(Y)) => eq(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) min(cons(0, nil)) => 0 min(cons(s(X), nil)) => s(X) min(cons(X, cons(Y, Z))) => ifmin(le(X, Y), cons(X, cons(Y, Z))) ifmin(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) ifmin(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) replace(X, Y, nil) => nil replace(X, Y, cons(Z, U)) => ifrepl(eq(X, Z), X, Y, cons(Z, U)) ifrepl(true, X, Y, cons(Z, U)) => cons(Y, U) ifrepl(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: selsort#(cons(X, Y)) >? ifselsort#(eq(X, min(cons(X, Y))), cons(X, Y)) ifselsort#(true, cons(X, Y)) >? selsort#(Y) ifselsort#(false, cons(X, Y)) >? selsort#(replace(min(cons(X, Y)), X, Y)) eq(0, 0) >= true eq(0, s(X)) >= false eq(s(X), 0) >= false eq(s(X), s(Y)) >= eq(X, Y) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) min(cons(0, nil)) >= 0 min(cons(s(X), nil)) >= s(X) min(cons(X, cons(Y, Z))) >= ifmin(le(X, Y), cons(X, cons(Y, Z))) ifmin(true, cons(X, cons(Y, Z))) >= min(cons(X, Z)) ifmin(false, cons(X, cons(Y, Z))) >= min(cons(Y, Z)) replace(X, Y, nil) >= nil replace(X, Y, cons(Z, U)) >= ifrepl(eq(X, Z), X, Y, cons(Z, U)) ifrepl(true, X, Y, cons(Z, U)) >= cons(Y, U) ifrepl(false, X, Y, cons(Z, U)) >= cons(Z, replace(X, Y, U)) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: ifrepl(x_1,x_2,x_3,x_4) = ifrepl(x_2x_3,x_4) ifselsort#(x_1,x_2) = ifselsort#(x_2) replace(x_1,x_2,x_3) = replace(x_2x_3) This leaves the following ordering requirements: selsort#(cons(X, Y)) >= ifselsort#(eq(X, min(cons(X, Y))), cons(X, Y)) ifselsort#(true, cons(X, Y)) >= selsort#(Y) ifselsort#(false, cons(X, Y)) > selsort#(replace(min(cons(X, Y)), X, Y)) replace(X, Y, nil) >= nil replace(X, Y, cons(Z, U)) >= ifrepl(eq(X, Z), X, Y, cons(Z, U)) ifrepl(true, X, Y, cons(Z, U)) >= cons(Y, U) ifrepl(false, X, Y, cons(Z, U)) >= cons(Z, replace(X, Y, U)) The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.1 + y1 eq = \y0y1.0 false = 0 ifmin = \y0y1.0 ifrepl = \y0y1y2y3.y3 ifselsort# = \y0y1.y1 le = \y0y1.0 min = \y0.0 nil = 0 replace = \y0y1y2.y2 s = \y0.0 selsort# = \y0.y0 true = 0 Using this interpretation, the requirements translate to: [[selsort#(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[ifselsort#(eq(_x0, min(cons(_x0, _x1))), cons(_x0, _x1))]] [[ifselsort#(true, cons(_x0, _x1))]] = 1 + x1 > x1 = [[selsort#(_x1)]] [[ifselsort#(false, cons(_x0, _x1))]] = 1 + x1 > x1 = [[selsort#(replace(min(cons(_x0, _x1)), _x0, _x1))]] [[replace(_x0, _x1, nil)]] = 0 >= 0 = [[nil]] [[replace(_x0, _x1, cons(_x2, _x3))]] = 1 + x3 >= 1 + x3 = [[ifrepl(eq(_x0, _x2), _x0, _x1, cons(_x2, _x3))]] [[ifrepl(true, _x0, _x1, cons(_x2, _x3))]] = 1 + x3 >= 1 + x3 = [[cons(_x1, _x3)]] [[ifrepl(false, _x0, _x1, cons(_x2, _x3))]] = 1 + x3 >= 1 + x3 = [[cons(_x2, replace(_x0, _x1, _x3))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_0, minimal, formative) by (P_6, R_0, minimal, formative), where P_6 consists of: selsort#(cons(X, Y)) =#> ifselsort#(eq(X, min(cons(X, Y))), cons(X, Y)) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_6, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_6, 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), (P_2, R_0, minimal, formative), (P_3, 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 apply the subterm criterion with the following projection function: nu(ifrepl#) = 4 nu(replace#) = 3 Thus, we can orient the dependency pairs as follows: nu(replace#(X, Y, cons(Z, U))) = cons(Z, U) = cons(Z, U) = nu(ifrepl#(eq(X, Z), X, Y, cons(Z, U))) nu(ifrepl#(false, X, Y, cons(Z, U))) = cons(Z, U) |> U = nu(replace#(X, Y, U)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_0, minimal, f) by (P_7, R_0, minimal, f), where P_7 contains: replace#(X, Y, cons(Z, U)) =#> ifrepl#(eq(X, Z), X, Y, cons(Z, U)) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_7, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_7, 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), (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) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: min#(cons(X, cons(Y, Z))) >? ifmin#(le(X, Y), cons(X, cons(Y, Z))) ifmin#(true, cons(X, cons(Y, Z))) >? min#(cons(X, Z)) ifmin#(false, cons(X, cons(Y, Z))) >? min#(cons(Y, Z)) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(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: ifmin#(x_1,x_2) = ifmin#(x_2) This leaves the following ordering requirements: min#(cons(X, cons(Y, Z))) > ifmin#(le(X, Y), cons(X, cons(Y, Z))) ifmin#(true, cons(X, cons(Y, Z))) >= min#(cons(X, Z)) ifmin#(false, cons(X, cons(Y, Z))) >= min#(cons(Y, Z)) The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.2 + 3y0 + 3y1 false = 0 ifmin# = \y0y1.y1 le = \y0y1.0 min# = \y0.2 + y0 s = \y0.3 true = 0 Using this interpretation, the requirements translate to: [[min#(cons(_x0, cons(_x1, _x2)))]] = 10 + 3x0 + 9x1 + 9x2 > 8 + 3x0 + 9x1 + 9x2 = [[ifmin#(le(_x0, _x1), cons(_x0, cons(_x1, _x2)))]] [[ifmin#(true, cons(_x0, cons(_x1, _x2)))]] = 8 + 3x0 + 9x1 + 9x2 > 4 + 3x0 + 3x2 = [[min#(cons(_x0, _x2))]] [[ifmin#(false, cons(_x0, cons(_x1, _x2)))]] = 8 + 3x0 + 9x1 + 9x2 > 4 + 3x1 + 3x2 = [[min#(cons(_x1, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, 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 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(eq#) = 1 Thus, we can orient the dependency pairs as follows: nu(eq#(s(X), s(Y))) = s(X) |> X = nu(eq#(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.