/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 add : [o * o] --> o app : [o * o] --> o false : [] --> o high : [o * o] --> o if!6220high : [o * o * o] --> o if!6220low : [o * o * o] --> o le : [o * o] --> o low : [o * o] --> o minus : [o * o] --> o nil : [] --> o quicksort : [o] --> o quot : [o * o] --> o s : [o] --> o true : [] --> o minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) app(nil, X) => X app(add(X, Y), Z) => add(X, app(Y, Z)) low(X, nil) => nil low(X, add(Y, Z)) => if!6220low(le(Y, X), X, add(Y, Z)) if!6220low(true, X, add(Y, Z)) => add(Y, low(X, Z)) if!6220low(false, X, add(Y, Z)) => low(X, Z) high(X, nil) => nil high(X, add(Y, Z)) => if!6220high(le(Y, X), X, add(Y, Z)) if!6220high(true, X, add(Y, Z)) => high(X, Z) if!6220high(false, X, add(Y, Z)) => add(Y, high(X, Z)) quicksort(nil) => nil quicksort(add(X, Y)) => app(quicksort(low(X, Y)), add(X, quicksort(high(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 : [] --> zg add : [zg * zg] --> zg app : [zg * zg] --> zg false : [] --> df high : [zg * zg] --> zg if!6220high : [df * zg * zg] --> zg if!6220low : [df * zg * zg] --> zg le : [zg * zg] --> df low : [zg * zg] --> zg minus : [zg * zg] --> zg nil : [] --> zg quicksort : [zg] --> zg quot : [zg * zg] --> zg s : [zg] --> zg true : [] --> df 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#(s(X), s(Y)) =#> minus#(X, Y) 1] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 2] quot#(s(X), s(Y)) =#> minus#(X, Y) 3] le#(s(X), s(Y)) =#> le#(X, Y) 4] app#(add(X, Y), Z) =#> app#(Y, Z) 5] low#(X, add(Y, Z)) =#> if!6220low#(le(Y, X), X, add(Y, Z)) 6] low#(X, add(Y, Z)) =#> le#(Y, X) 7] if!6220low#(true, X, add(Y, Z)) =#> low#(X, Z) 8] if!6220low#(false, X, add(Y, Z)) =#> low#(X, Z) 9] high#(X, add(Y, Z)) =#> if!6220high#(le(Y, X), X, add(Y, Z)) 10] high#(X, add(Y, Z)) =#> le#(Y, X) 11] if!6220high#(true, X, add(Y, Z)) =#> high#(X, Z) 12] if!6220high#(false, X, add(Y, Z)) =#> high#(X, Z) 13] quicksort#(add(X, Y)) =#> app#(quicksort(low(X, Y)), add(X, quicksort(high(X, Y)))) 14] quicksort#(add(X, Y)) =#> quicksort#(low(X, Y)) 15] quicksort#(add(X, Y)) =#> low#(X, Y) 16] quicksort#(add(X, Y)) =#> quicksort#(high(X, Y)) 17] quicksort#(add(X, Y)) =#> high#(X, Y) Rules R_0: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) app(nil, X) => X app(add(X, Y), Z) => add(X, app(Y, Z)) low(X, nil) => nil low(X, add(Y, Z)) => if!6220low(le(Y, X), X, add(Y, Z)) if!6220low(true, X, add(Y, Z)) => add(Y, low(X, Z)) if!6220low(false, X, add(Y, Z)) => low(X, Z) high(X, nil) => nil high(X, add(Y, Z)) => if!6220high(le(Y, X), X, add(Y, Z)) if!6220high(true, X, add(Y, Z)) => high(X, Z) if!6220high(false, X, add(Y, Z)) => add(Y, high(X, Z)) quicksort(nil) => nil quicksort(add(X, Y)) => app(quicksort(low(X, Y)), add(X, quicksort(high(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 * 2 : 0 * 3 : 3 * 4 : 4 * 5 : 7, 8 * 6 : 3 * 7 : 5, 6 * 8 : 5, 6 * 9 : 11, 12 * 10 : 3 * 11 : 9, 10 * 12 : 9, 10 * 13 : 4 * 14 : 13, 14, 15, 16, 17 * 15 : 5, 6 * 16 : 13, 14, 15, 16, 17 * 17 : 9, 10 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) P_2: quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) P_3: le#(s(X), s(Y)) =#> le#(X, Y) P_4: app#(add(X, Y), Z) =#> app#(Y, Z) P_5: low#(X, add(Y, Z)) =#> if!6220low#(le(Y, X), X, add(Y, Z)) if!6220low#(true, X, add(Y, Z)) =#> low#(X, Z) if!6220low#(false, X, add(Y, Z)) =#> low#(X, Z) P_6: high#(X, add(Y, Z)) =#> if!6220high#(le(Y, X), X, add(Y, Z)) if!6220high#(true, X, add(Y, Z)) =#> high#(X, Z) if!6220high#(false, X, add(Y, Z)) =#> high#(X, Z) P_7: quicksort#(add(X, Y)) =#> quicksort#(low(X, Y)) quicksort#(add(X, Y)) =#> quicksort#(high(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), (P_5, R_0, m, f), (P_6, R_0, m, f) and (P_7, 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), (P_5, R_0, minimal, formative), (P_6, 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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_7, R_0) are: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) low(X, nil) => nil low(X, add(Y, Z)) => if!6220low(le(Y, X), X, add(Y, Z)) if!6220low(true, X, add(Y, Z)) => add(Y, low(X, Z)) if!6220low(false, X, add(Y, Z)) => low(X, Z) high(X, nil) => nil high(X, add(Y, Z)) => if!6220high(le(Y, X), X, add(Y, Z)) if!6220high(true, X, add(Y, Z)) => high(X, Z) if!6220high(false, X, add(Y, Z)) => add(Y, high(X, Z)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: quicksort#(add(X, Y)) >? quicksort#(low(X, Y)) quicksort#(add(X, Y)) >? quicksort#(high(X, Y)) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) low(X, nil) >= nil low(X, add(Y, Z)) >= if!6220low(le(Y, X), X, add(Y, Z)) if!6220low(true, X, add(Y, Z)) >= add(Y, low(X, Z)) if!6220low(false, X, add(Y, Z)) >= low(X, Z) high(X, nil) >= nil high(X, add(Y, Z)) >= if!6220high(le(Y, X), X, add(Y, Z)) if!6220high(true, X, add(Y, Z)) >= high(X, Z) if!6220high(false, X, add(Y, Z)) >= add(Y, high(X, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: if!6220high(x_1,x_2,x_3) = if!6220high(x_2x_3) if!6220low(x_1,x_2,x_3) = if!6220low(x_2x_3) This leaves the following ordering requirements: quicksort#(add(X, Y)) > quicksort#(low(X, Y)) quicksort#(add(X, Y)) >= quicksort#(high(X, Y)) low(X, nil) >= nil low(X, add(Y, Z)) >= if!6220low(le(Y, X), X, add(Y, Z)) if!6220low(true, X, add(Y, Z)) >= add(Y, low(X, Z)) if!6220low(false, X, add(Y, Z)) >= low(X, Z) high(X, nil) >= nil high(X, add(Y, Z)) >= if!6220high(le(Y, X), X, add(Y, Z)) if!6220high(true, X, add(Y, Z)) >= high(X, Z) if!6220high(false, X, add(Y, Z)) >= add(Y, high(X, Z)) The following interpretation satisfies the requirements: 0 = 3 add = \y0y1.3 + y1 false = 0 high = \y0y1.y1 if!6220high = \y0y1y2.y2 if!6220low = \y0y1y2.y2 le = \y0y1.0 low = \y0y1.y1 nil = 0 quicksort# = \y0.3y0 s = \y0.3 true = 0 Using this interpretation, the requirements translate to: [[quicksort#(add(_x0, _x1))]] = 9 + 3x1 > 3x1 = [[quicksort#(low(_x0, _x1))]] [[quicksort#(add(_x0, _x1))]] = 9 + 3x1 > 3x1 = [[quicksort#(high(_x0, _x1))]] [[low(_x0, nil)]] = 0 >= 0 = [[nil]] [[low(_x0, add(_x1, _x2))]] = 3 + x2 >= 3 + x2 = [[if!6220low(le(_x1, _x0), _x0, add(_x1, _x2))]] [[if!6220low(true, _x0, add(_x1, _x2))]] = 3 + x2 >= 3 + x2 = [[add(_x1, low(_x0, _x2))]] [[if!6220low(false, _x0, add(_x1, _x2))]] = 3 + x2 >= x2 = [[low(_x0, _x2)]] [[high(_x0, nil)]] = 0 >= 0 = [[nil]] [[high(_x0, add(_x1, _x2))]] = 3 + x2 >= 3 + x2 = [[if!6220high(le(_x1, _x0), _x0, add(_x1, _x2))]] [[if!6220high(true, _x0, add(_x1, _x2))]] = 3 + x2 >= x2 = [[high(_x0, _x2)]] [[if!6220high(false, _x0, add(_x1, _x2))]] = 3 + x2 >= 3 + x2 = [[add(_x1, high(_x0, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_7, 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), (P_4, R_0, minimal, formative), (P_5, 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 apply the subterm criterion with the following projection function: nu(high#) = 2 nu(if!6220high#) = 3 Thus, we can orient the dependency pairs as follows: nu(high#(X, add(Y, Z))) = add(Y, Z) = add(Y, Z) = nu(if!6220high#(le(Y, X), X, add(Y, Z))) nu(if!6220high#(true, X, add(Y, Z))) = add(Y, Z) |> Z = nu(high#(X, Z)) nu(if!6220high#(false, X, add(Y, Z))) = add(Y, Z) |> Z = nu(high#(X, Z)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_6, R_0, minimal, f) by (P_8, R_0, minimal, f), where P_8 contains: high#(X, add(Y, Z)) =#> if!6220high#(le(Y, X), X, add(Y, Z)) 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), (P_5, R_0, minimal, formative) and (P_8, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_8, 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), (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 apply the subterm criterion with the following projection function: nu(if!6220low#) = 3 nu(low#) = 2 Thus, we can orient the dependency pairs as follows: nu(low#(X, add(Y, Z))) = add(Y, Z) = add(Y, Z) = nu(if!6220low#(le(Y, X), X, add(Y, Z))) nu(if!6220low#(true, X, add(Y, Z))) = add(Y, Z) |> Z = nu(low#(X, Z)) nu(if!6220low#(false, X, add(Y, Z))) = add(Y, Z) |> Z = nu(low#(X, Z)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_0, minimal, f) by (P_9, R_0, minimal, f), where P_9 contains: low#(X, add(Y, Z)) =#> if!6220low#(le(Y, X), X, add(Y, Z)) 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_9, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_9, 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(app#) = 1 Thus, we can orient the dependency pairs as follows: nu(app#(add(X, Y), Z)) = add(X, Y) |> Y = nu(app#(Y, Z)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, 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 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 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_3, 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 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: minus(X, 0) => X minus(s(X), s(Y)) => 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)) minus(X, 0) >= X minus(s(X), s(Y)) >= 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.y0 quot# = \y0y1.3y0 s = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[quot#(s(_x0), s(_x1))]] = 9 + 9x0 > 3x0 = [[quot#(minus(_x0, _x1), s(_x1))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 3 + 3x0 >= x0 = [[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#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), s(Y))) = s(X) |> X = 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.