/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 div : [o * o] --> o div!6220active : [o * o] --> o false : [] --> o ge : [o * o] --> o ge!6220active : [o * o] --> o if : [o * o * o] --> o if!6220active : [o * o * o] --> o mark : [o] --> o minus : [o * o] --> o minus!6220active : [o * o] --> o s : [o] --> o true : [] --> o minus!6220active(0, X) => 0 mark(0) => 0 minus!6220active(s(X), s(Y)) => minus!6220active(X, Y) mark(s(X)) => s(mark(X)) ge!6220active(X, 0) => true mark(minus(X, Y)) => minus!6220active(X, Y) ge!6220active(0, s(X)) => false mark(ge(X, Y)) => ge!6220active(X, Y) ge!6220active(s(X), s(Y)) => ge!6220active(X, Y) mark(div(X, Y)) => div!6220active(mark(X), Y) div!6220active(0, s(X)) => 0 mark(if(X, Y, Z)) => if!6220active(mark(X), Y, Z) div!6220active(s(X), s(Y)) => if!6220active(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active(true, X, Y) => mark(X) minus!6220active(X, Y) => minus(X, Y) if!6220active(false, X, Y) => mark(Y) ge!6220active(X, Y) => ge(X, Y) if!6220active(X, Y, Z) => if(X, Y, Z) div!6220active(X, Y) => div(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, formative): Dependency Pairs P_0: 0] minus!6220active#(s(X), s(Y)) =#> minus!6220active#(X, Y) 1] mark#(s(X)) =#> mark#(X) 2] mark#(minus(X, Y)) =#> minus!6220active#(X, Y) 3] mark#(ge(X, Y)) =#> ge!6220active#(X, Y) 4] ge!6220active#(s(X), s(Y)) =#> ge!6220active#(X, Y) 5] mark#(div(X, Y)) =#> div!6220active#(mark(X), Y) 6] mark#(div(X, Y)) =#> mark#(X) 7] mark#(if(X, Y, Z)) =#> if!6220active#(mark(X), Y, Z) 8] mark#(if(X, Y, Z)) =#> mark#(X) 9] div!6220active#(s(X), s(Y)) =#> if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) 10] div!6220active#(s(X), s(Y)) =#> ge!6220active#(X, Y) 11] if!6220active#(true, X, Y) =#> mark#(X) 12] if!6220active#(false, X, Y) =#> mark#(Y) Rules R_0: minus!6220active(0, X) => 0 mark(0) => 0 minus!6220active(s(X), s(Y)) => minus!6220active(X, Y) mark(s(X)) => s(mark(X)) ge!6220active(X, 0) => true mark(minus(X, Y)) => minus!6220active(X, Y) ge!6220active(0, s(X)) => false mark(ge(X, Y)) => ge!6220active(X, Y) ge!6220active(s(X), s(Y)) => ge!6220active(X, Y) mark(div(X, Y)) => div!6220active(mark(X), Y) div!6220active(0, s(X)) => 0 mark(if(X, Y, Z)) => if!6220active(mark(X), Y, Z) div!6220active(s(X), s(Y)) => if!6220active(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active(true, X, Y) => mark(X) minus!6220active(X, Y) => minus(X, Y) if!6220active(false, X, Y) => mark(Y) ge!6220active(X, Y) => ge(X, Y) if!6220active(X, Y, Z) => if(X, Y, Z) div!6220active(X, Y) => div(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, 3, 5, 6, 7, 8 * 2 : 0 * 3 : 4 * 4 : 4 * 5 : 9, 10 * 6 : 1, 2, 3, 5, 6, 7, 8 * 7 : 11, 12 * 8 : 1, 2, 3, 5, 6, 7, 8 * 9 : 11, 12 * 10 : 4 * 11 : 1, 2, 3, 5, 6, 7, 8 * 12 : 1, 2, 3, 5, 6, 7, 8 This graph has the following strongly connected components: P_1: minus!6220active#(s(X), s(Y)) =#> minus!6220active#(X, Y) P_2: mark#(s(X)) =#> mark#(X) mark#(div(X, Y)) =#> div!6220active#(mark(X), Y) mark#(div(X, Y)) =#> mark#(X) mark#(if(X, Y, Z)) =#> if!6220active#(mark(X), Y, Z) mark#(if(X, Y, Z)) =#> mark#(X) div!6220active#(s(X), s(Y)) =#> if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active#(true, X, Y) =#> mark#(X) if!6220active#(false, X, Y) =#> mark#(Y) P_3: ge!6220active#(s(X), s(Y)) =#> ge!6220active#(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) 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 apply the subterm criterion with the following projection function: nu(ge!6220active#) = 1 Thus, we can orient the dependency pairs as follows: nu(ge!6220active#(s(X), s(Y))) = s(X) |> X = nu(ge!6220active#(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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: mark#(s(X)) >? mark#(X) mark#(div(X, Y)) >? div!6220active#(mark(X), Y) mark#(div(X, Y)) >? mark#(X) mark#(if(X, Y, Z)) >? if!6220active#(mark(X), Y, Z) mark#(if(X, Y, Z)) >? mark#(X) div!6220active#(s(X), s(Y)) >? if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active#(true, X, Y) >? mark#(X) if!6220active#(false, X, Y) >? mark#(Y) minus!6220active(0, X) >= 0 mark(0) >= 0 minus!6220active(s(X), s(Y)) >= minus!6220active(X, Y) mark(s(X)) >= s(mark(X)) ge!6220active(X, 0) >= true mark(minus(X, Y)) >= minus!6220active(X, Y) ge!6220active(0, s(X)) >= false mark(ge(X, Y)) >= ge!6220active(X, Y) ge!6220active(s(X), s(Y)) >= ge!6220active(X, Y) mark(div(X, Y)) >= div!6220active(mark(X), Y) div!6220active(0, s(X)) >= 0 mark(if(X, Y, Z)) >= if!6220active(mark(X), Y, Z) div!6220active(s(X), s(Y)) >= if!6220active(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active(true, X, Y) >= mark(X) minus!6220active(X, Y) >= minus(X, Y) if!6220active(false, X, Y) >= mark(Y) ge!6220active(X, Y) >= ge(X, Y) if!6220active(X, Y, Z) >= if(X, Y, Z) div!6220active(X, Y) >= div(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: div!6220active#(x_1,x_2) = div!6220active#(x_2) if!6220active#(x_1,x_2,x_3) = if!6220active#(x_2x_3) This leaves the following ordering requirements: mark#(s(X)) >= mark#(X) mark#(div(X, Y)) >= div!6220active#(mark(X), Y) mark#(div(X, Y)) >= mark#(X) mark#(if(X, Y, Z)) >= if!6220active#(mark(X), Y, Z) mark#(if(X, Y, Z)) > mark#(X) div!6220active#(s(X), s(Y)) >= if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active#(true, X, Y) >= mark#(X) if!6220active#(false, X, Y) >= mark#(Y) The following interpretation satisfies the requirements: 0 = 0 div = \y0y1.2y0 div!6220active = \y0y1.0 div!6220active# = \y0y1.0 false = 0 ge = \y0y1.0 ge!6220active = \y0y1.0 if = \y0y1y2.1 + 2y0 + 2y1 + 2y2 if!6220active = \y0y1y2.0 if!6220active# = \y0y1y2.y1 + y2 mark = \y0.0 mark# = \y0.y0 minus = \y0y1.0 minus!6220active = \y0y1.0 s = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[mark#(s(_x0))]] = 2x0 >= x0 = [[mark#(_x0)]] [[mark#(div(_x0, _x1))]] = 2x0 >= 0 = [[div!6220active#(mark(_x0), _x1)]] [[mark#(div(_x0, _x1))]] = 2x0 >= x0 = [[mark#(_x0)]] [[mark#(if(_x0, _x1, _x2))]] = 1 + 2x0 + 2x1 + 2x2 > x1 + x2 = [[if!6220active#(mark(_x0), _x1, _x2)]] [[mark#(if(_x0, _x1, _x2))]] = 1 + 2x0 + 2x1 + 2x2 > x0 = [[mark#(_x0)]] [[div!6220active#(s(_x0), s(_x1))]] = 0 >= 0 = [[if!6220active#(ge!6220active(_x0, _x1), s(div(minus(_x0, _x1), s(_x1))), 0)]] [[if!6220active#(true, _x0, _x1)]] = x0 + x1 >= x0 = [[mark#(_x0)]] [[if!6220active#(false, _x0, _x1)]] = x0 + x1 >= x1 = [[mark#(_x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_4, R_0, minimal, formative), where P_4 consists of: mark#(s(X)) =#> mark#(X) mark#(div(X, Y)) =#> div!6220active#(mark(X), Y) mark#(div(X, Y)) =#> mark#(X) div!6220active#(s(X), s(Y)) =#> if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active#(true, X, Y) =#> mark#(X) if!6220active#(false, X, Y) =#> mark#(Y) Thus, the original system is terminating if each of (P_1, 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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: mark#(s(X)) >? mark#(X) mark#(div(X, Y)) >? div!6220active#(mark(X), Y) mark#(div(X, Y)) >? mark#(X) div!6220active#(s(X), s(Y)) >? if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active#(true, X, Y) >? mark#(X) if!6220active#(false, X, Y) >? mark#(Y) minus!6220active(0, X) >= 0 mark(0) >= 0 minus!6220active(s(X), s(Y)) >= minus!6220active(X, Y) mark(s(X)) >= s(mark(X)) ge!6220active(X, 0) >= true mark(minus(X, Y)) >= minus!6220active(X, Y) ge!6220active(0, s(X)) >= false mark(ge(X, Y)) >= ge!6220active(X, Y) ge!6220active(s(X), s(Y)) >= ge!6220active(X, Y) mark(div(X, Y)) >= div!6220active(mark(X), Y) div!6220active(0, s(X)) >= 0 mark(if(X, Y, Z)) >= if!6220active(mark(X), Y, Z) div!6220active(s(X), s(Y)) >= if!6220active(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active(true, X, Y) >= mark(X) minus!6220active(X, Y) >= minus(X, Y) if!6220active(false, X, Y) >= mark(Y) ge!6220active(X, Y) >= ge(X, Y) if!6220active(X, Y, Z) >= if(X, Y, Z) div!6220active(X, Y) >= div(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: div!6220active#(x_1,x_2) = div!6220active#(x_2) if!6220active#(x_1,x_2,x_3) = if!6220active#(x_2x_3) This leaves the following ordering requirements: mark#(s(X)) >= mark#(X) mark#(div(X, Y)) >= div!6220active#(mark(X), Y) mark#(div(X, Y)) > mark#(X) div!6220active#(s(X), s(Y)) >= if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active#(true, X, Y) >= mark#(X) if!6220active#(false, X, Y) >= mark#(Y) The following interpretation satisfies the requirements: 0 = 0 div = \y0y1.1 + y0 div!6220active = \y0y1.0 div!6220active# = \y0y1.2 false = 0 ge = \y0y1.0 ge!6220active = \y0y1.0 if = \y0y1y2.0 if!6220active = \y0y1y2.0 if!6220active# = \y0y1y2.2y1 + 2y2 mark = \y0.0 mark# = \y0.2y0 minus = \y0y1.0 minus!6220active = \y0y1.0 s = \y0.y0 true = 0 Using this interpretation, the requirements translate to: [[mark#(s(_x0))]] = 2x0 >= 2x0 = [[mark#(_x0)]] [[mark#(div(_x0, _x1))]] = 2 + 2x0 >= 2 = [[div!6220active#(mark(_x0), _x1)]] [[mark#(div(_x0, _x1))]] = 2 + 2x0 > 2x0 = [[mark#(_x0)]] [[div!6220active#(s(_x0), s(_x1))]] = 2 >= 2 = [[if!6220active#(ge!6220active(_x0, _x1), s(div(minus(_x0, _x1), s(_x1))), 0)]] [[if!6220active#(true, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 = [[mark#(_x0)]] [[if!6220active#(false, _x0, _x1)]] = 2x0 + 2x1 >= 2x1 = [[mark#(_x1)]] 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: mark#(s(X)) =#> mark#(X) mark#(div(X, Y)) =#> div!6220active#(mark(X), Y) div!6220active#(s(X), s(Y)) =#> if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active#(true, X, Y) =#> mark#(X) if!6220active#(false, X, Y) =#> mark#(Y) Thus, the original system is terminating if each of (P_1, 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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: mark#(s(X)) >? mark#(X) mark#(div(X, Y)) >? div!6220active#(mark(X), Y) div!6220active#(s(X), s(Y)) >? if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active#(true, X, Y) >? mark#(X) if!6220active#(false, X, Y) >? mark#(Y) minus!6220active(0, X) >= 0 mark(0) >= 0 minus!6220active(s(X), s(Y)) >= minus!6220active(X, Y) mark(s(X)) >= s(mark(X)) ge!6220active(X, 0) >= true mark(minus(X, Y)) >= minus!6220active(X, Y) ge!6220active(0, s(X)) >= false mark(ge(X, Y)) >= ge!6220active(X, Y) ge!6220active(s(X), s(Y)) >= ge!6220active(X, Y) mark(div(X, Y)) >= div!6220active(mark(X), Y) div!6220active(0, s(X)) >= 0 mark(if(X, Y, Z)) >= if!6220active(mark(X), Y, Z) div!6220active(s(X), s(Y)) >= if!6220active(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active(true, X, Y) >= mark(X) minus!6220active(X, Y) >= minus(X, Y) if!6220active(false, X, Y) >= mark(Y) ge!6220active(X, Y) >= ge(X, Y) if!6220active(X, Y, Z) >= if(X, Y, Z) div!6220active(X, Y) >= div(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 div = \y0y1.2 + 3y0 div!6220active = \y0y1.2 + 3y0 div!6220active# = \y0y1.1 + 3y0 false = 0 ge = \y0y1.2y1 ge!6220active = \y0y1.2y1 if = \y0y1y2.y1 + y2 if!6220active = \y0y1y2.y1 + y2 if!6220active# = \y0y1y2.y1 + 2y2 mark = \y0.y0 mark# = \y0.y0 minus = \y0y1.0 minus!6220active = \y0y1.0 s = \y0.2 + 2y0 true = 0 Using this interpretation, the requirements translate to: [[mark#(s(_x0))]] = 2 + 2x0 > x0 = [[mark#(_x0)]] [[mark#(div(_x0, _x1))]] = 2 + 3x0 > 1 + 3x0 = [[div!6220active#(mark(_x0), _x1)]] [[div!6220active#(s(_x0), s(_x1))]] = 7 + 6x0 > 6 = [[if!6220active#(ge!6220active(_x0, _x1), s(div(minus(_x0, _x1), s(_x1))), 0)]] [[if!6220active#(true, _x0, _x1)]] = x0 + 2x1 >= x0 = [[mark#(_x0)]] [[if!6220active#(false, _x0, _x1)]] = x0 + 2x1 >= x1 = [[mark#(_x1)]] [[minus!6220active(0, _x0)]] = 0 >= 0 = [[0]] [[mark(0)]] = 0 >= 0 = [[0]] [[minus!6220active(s(_x0), s(_x1))]] = 0 >= 0 = [[minus!6220active(_x0, _x1)]] [[mark(s(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[s(mark(_x0))]] [[ge!6220active(_x0, 0)]] = 0 >= 0 = [[true]] [[mark(minus(_x0, _x1))]] = 0 >= 0 = [[minus!6220active(_x0, _x1)]] [[ge!6220active(0, s(_x0))]] = 4 + 4x0 >= 0 = [[false]] [[mark(ge(_x0, _x1))]] = 2x1 >= 2x1 = [[ge!6220active(_x0, _x1)]] [[ge!6220active(s(_x0), s(_x1))]] = 4 + 4x1 >= 2x1 = [[ge!6220active(_x0, _x1)]] [[mark(div(_x0, _x1))]] = 2 + 3x0 >= 2 + 3x0 = [[div!6220active(mark(_x0), _x1)]] [[div!6220active(0, s(_x0))]] = 2 >= 0 = [[0]] [[mark(if(_x0, _x1, _x2))]] = x1 + x2 >= x1 + x2 = [[if!6220active(mark(_x0), _x1, _x2)]] [[div!6220active(s(_x0), s(_x1))]] = 8 + 6x0 >= 6 = [[if!6220active(ge!6220active(_x0, _x1), s(div(minus(_x0, _x1), s(_x1))), 0)]] [[if!6220active(true, _x0, _x1)]] = x0 + x1 >= x0 = [[mark(_x0)]] [[minus!6220active(_x0, _x1)]] = 0 >= 0 = [[minus(_x0, _x1)]] [[if!6220active(false, _x0, _x1)]] = x0 + x1 >= x1 = [[mark(_x1)]] [[ge!6220active(_x0, _x1)]] = 2x1 >= 2x1 = [[ge(_x0, _x1)]] [[if!6220active(_x0, _x1, _x2)]] = x1 + x2 >= x1 + x2 = [[if(_x0, _x1, _x2)]] [[div!6220active(_x0, _x1)]] = 2 + 3x0 >= 2 + 3x0 = [[div(_x0, _x1)]] 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: if!6220active#(true, X, Y) =#> mark#(X) if!6220active#(false, X, Y) =#> mark#(Y) Thus, the original system is terminating if each of (P_1, 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 : * 1 : 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 (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!6220active#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus!6220active#(s(X), s(Y))) = s(X) |> X = nu(minus!6220active#(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 +++ [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.