/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 if!6220min : [o * o] --> o if!6220replace : [o * o * o * o] --> o le : [o * o] --> o min : [o] --> o nil : [] --> o replace : [o * o * o] --> o s : [o] --> o sort : [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))) => if!6220min(le(X, Y), cons(X, cons(Y, Z))) if!6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) if!6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) replace(X, Y, nil) => nil replace(X, Y, cons(Z, U)) => if!6220replace(eq(X, Z), X, Y, cons(Z, U)) if!6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) if!6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) sort(nil) => nil sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(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 : [] --> pg cons : [pg * ug] --> ug eq : [pg * pg] --> se false : [] --> se if!6220min : [se * ug] --> pg if!6220replace : [se * pg * pg * ug] --> ug le : [pg * pg] --> se min : [ug] --> pg nil : [] --> ug replace : [pg * pg * ug] --> ug s : [pg] --> pg sort : [ug] --> ug true : [] --> se 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))) =#> if!6220min#(le(X, Y), cons(X, cons(Y, Z))) 3] min#(cons(X, cons(Y, Z))) =#> le#(X, Y) 4] if!6220min#(true, cons(X, cons(Y, Z))) =#> min#(cons(X, Z)) 5] if!6220min#(false, cons(X, cons(Y, Z))) =#> min#(cons(Y, Z)) 6] replace#(X, Y, cons(Z, U)) =#> if!6220replace#(eq(X, Z), X, Y, cons(Z, U)) 7] replace#(X, Y, cons(Z, U)) =#> eq#(X, Z) 8] if!6220replace#(false, X, Y, cons(Z, U)) =#> replace#(X, Y, U) 9] sort#(cons(X, Y)) =#> min#(cons(X, Y)) 10] sort#(cons(X, Y)) =#> sort#(replace(min(cons(X, Y)), X, Y)) 11] sort#(cons(X, Y)) =#> replace#(min(cons(X, Y)), X, Y) 12] sort#(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))) => if!6220min(le(X, Y), cons(X, cons(Y, Z))) if!6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) if!6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) replace(X, Y, nil) => nil replace(X, Y, cons(Z, U)) => if!6220replace(eq(X, Z), X, Y, cons(Z, U)) if!6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) if!6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) sort(nil) => nil sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(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 : 2, 3 * 10 : 9, 10, 11, 12 * 11 : 6, 7 * 12 : 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))) =#> if!6220min#(le(X, Y), cons(X, cons(Y, Z))) if!6220min#(true, cons(X, cons(Y, Z))) =#> min#(cons(X, Z)) if!6220min#(false, cons(X, cons(Y, Z))) =#> min#(cons(Y, Z)) P_4: replace#(X, Y, cons(Z, U)) =#> if!6220replace#(eq(X, Z), X, Y, cons(Z, U)) if!6220replace#(false, X, Y, cons(Z, U)) =#> replace#(X, Y, U) P_5: sort#(cons(X, Y)) =#> sort#(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))) => if!6220min(le(X, Y), cons(X, cons(Y, Z))) if!6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) if!6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) replace(X, Y, nil) => nil replace(X, Y, cons(Z, U)) => if!6220replace(eq(X, Z), X, Y, cons(Z, U)) if!6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) if!6220replace(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: sort#(cons(X, Y)) >? sort#(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))) >= if!6220min(le(X, Y), cons(X, cons(Y, Z))) if!6220min(true, cons(X, cons(Y, Z))) >= min(cons(X, Z)) if!6220min(false, cons(X, cons(Y, Z))) >= min(cons(Y, Z)) replace(X, Y, nil) >= nil replace(X, Y, cons(Z, U)) >= if!6220replace(eq(X, Z), X, Y, cons(Z, U)) if!6220replace(true, X, Y, cons(Z, U)) >= cons(Y, U) if!6220replace(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: if!6220replace(x_1,x_2,x_3,x_4) = if!6220replace(x_2x_3,x_4) replace(x_1,x_2,x_3) = replace(x_2x_3) This leaves the following ordering requirements: sort#(cons(X, Y)) > sort#(replace(min(cons(X, Y)), X, Y)) replace(X, Y, nil) >= nil replace(X, Y, cons(Z, U)) >= if!6220replace(eq(X, Z), X, Y, cons(Z, U)) if!6220replace(true, X, Y, cons(Z, U)) >= cons(Y, U) if!6220replace(false, X, Y, cons(Z, U)) >= cons(Z, replace(X, Y, U)) The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.1 + 2y1 eq = \y0y1.0 false = 0 if!6220min = \y0y1.0 if!6220replace = \y0y1y2y3.y3 le = \y0y1.0 min = \y0.0 nil = 0 replace = \y0y1y2.y2 s = \y0.0 sort# = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[sort#(cons(_x0, _x1))]] = 2 + 4x1 > 2x1 = [[sort#(replace(min(cons(_x0, _x1)), _x0, _x1))]] [[replace(_x0, _x1, nil)]] = 0 >= 0 = [[nil]] [[replace(_x0, _x1, cons(_x2, _x3))]] = 1 + 2x3 >= 1 + 2x3 = [[if!6220replace(eq(_x0, _x2), _x0, _x1, cons(_x2, _x3))]] [[if!6220replace(true, _x0, _x1, cons(_x2, _x3))]] = 1 + 2x3 >= 1 + 2x3 = [[cons(_x1, _x3)]] [[if!6220replace(false, _x0, _x1, cons(_x2, _x3))]] = 1 + 2x3 >= 1 + 2x3 = [[cons(_x2, replace(_x0, _x1, _x3))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_5, 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), (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(if!6220replace#) = 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(if!6220replace#(eq(X, Z), X, Y, cons(Z, U))) nu(if!6220replace#(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_6, R_0, minimal, f), where P_6 contains: replace#(X, Y, cons(Z, U)) =#> if!6220replace#(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_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) 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))) >? if!6220min#(le(X, Y), cons(X, cons(Y, Z))) if!6220min#(true, cons(X, cons(Y, Z))) >? min#(cons(X, Z)) if!6220min#(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: if!6220min#(x_1,x_2) = if!6220min#(x_2) This leaves the following ordering requirements: min#(cons(X, cons(Y, Z))) > if!6220min#(le(X, Y), cons(X, cons(Y, Z))) if!6220min#(true, cons(X, cons(Y, Z))) >= min#(cons(X, Z)) if!6220min#(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 if!6220min# = \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 = [[if!6220min#(le(_x0, _x1), cons(_x0, cons(_x1, _x2)))]] [[if!6220min#(true, cons(_x0, cons(_x1, _x2)))]] = 8 + 3x0 + 9x1 + 9x2 > 4 + 3x0 + 3x2 = [[min#(cons(_x0, _x2))]] [[if!6220min#(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.