/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 a!6220!6220from : [o] --> o a!6220!6220length : [o] --> o a!6220!6220length1 : [o] --> o cons : [o * o] --> o from : [o] --> o length : [o] --> o length1 : [o] --> o mark : [o] --> o nil : [] --> o s : [o] --> o a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220length(nil) => 0 a!6220!6220length(cons(X, Y)) => s(a!6220!6220length1(Y)) a!6220!6220length1(X) => a!6220!6220length(X) mark(from(X)) => a!6220!6220from(mark(X)) mark(length(X)) => a!6220!6220length(X) mark(length1(X)) => a!6220!6220length1(X) mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) mark(nil) => nil mark(0) => 0 a!6220!6220from(X) => from(X) a!6220!6220length(X) => length(X) a!6220!6220length1(X) => length1(X) 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] a!6220!6220from#(X) =#> mark#(X) 1] a!6220!6220length#(cons(X, Y)) =#> a!6220!6220length1#(Y) 2] a!6220!6220length1#(X) =#> a!6220!6220length#(X) 3] mark#(from(X)) =#> a!6220!6220from#(mark(X)) 4] mark#(from(X)) =#> mark#(X) 5] mark#(length(X)) =#> a!6220!6220length#(X) 6] mark#(length1(X)) =#> a!6220!6220length1#(X) 7] mark#(cons(X, Y)) =#> mark#(X) 8] mark#(s(X)) =#> mark#(X) Rules R_0: a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220length(nil) => 0 a!6220!6220length(cons(X, Y)) => s(a!6220!6220length1(Y)) a!6220!6220length1(X) => a!6220!6220length(X) mark(from(X)) => a!6220!6220from(mark(X)) mark(length(X)) => a!6220!6220length(X) mark(length1(X)) => a!6220!6220length1(X) mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) mark(nil) => nil mark(0) => 0 a!6220!6220from(X) => from(X) a!6220!6220length(X) => length(X) a!6220!6220length1(X) => length1(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 : 3, 4, 5, 6, 7, 8 * 1 : 2 * 2 : 1 * 3 : 0 * 4 : 3, 4, 5, 6, 7, 8 * 5 : 1 * 6 : 2 * 7 : 3, 4, 5, 6, 7, 8 * 8 : 3, 4, 5, 6, 7, 8 This graph has the following strongly connected components: P_1: a!6220!6220from#(X) =#> mark#(X) mark#(from(X)) =#> a!6220!6220from#(mark(X)) mark#(from(X)) =#> mark#(X) mark#(cons(X, Y)) =#> mark#(X) mark#(s(X)) =#> mark#(X) P_2: a!6220!6220length#(cons(X, Y)) =#> a!6220!6220length1#(Y) a!6220!6220length1#(X) =#> a!6220!6220length#(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) and (P_2, R_0, m, f). 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(a!6220!6220length1#) = 1 nu(a!6220!6220length#) = 1 Thus, we can orient the dependency pairs as follows: nu(a!6220!6220length#(cons(X, Y))) = cons(X, Y) |> Y = nu(a!6220!6220length1#(Y)) nu(a!6220!6220length1#(X)) = X = X = nu(a!6220!6220length#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains: a!6220!6220length1#(X) =#> a!6220!6220length#(X) Thus, the original system is terminating if each of (P_1, 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 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 (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). The formative rules of (P_1, R_0) are R_1 ::= a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220length(cons(X, Y)) => s(a!6220!6220length1(Y)) a!6220!6220length1(X) => a!6220!6220length(X) mark(from(X)) => a!6220!6220from(mark(X)) mark(length(X)) => a!6220!6220length(X) mark(length1(X)) => a!6220!6220length1(X) mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) a!6220!6220from(X) => from(X) a!6220!6220length(X) => length(X) a!6220!6220length1(X) => length1(X) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative). Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_1, 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: a!6220!6220from#(X) >? mark#(X) mark#(from(X)) >? a!6220!6220from#(mark(X)) mark#(from(X)) >? mark#(X) mark#(cons(X, Y)) >? mark#(X) mark#(s(X)) >? mark#(X) a!6220!6220from(X) >= cons(mark(X), from(s(X))) a!6220!6220length(cons(X, Y)) >= s(a!6220!6220length1(Y)) a!6220!6220length1(X) >= a!6220!6220length(X) mark(from(X)) >= a!6220!6220from(mark(X)) mark(length(X)) >= a!6220!6220length(X) mark(length1(X)) >= a!6220!6220length1(X) mark(cons(X, Y)) >= cons(mark(X), Y) mark(s(X)) >= s(mark(X)) a!6220!6220from(X) >= from(X) a!6220!6220length(X) >= length(X) a!6220!6220length1(X) >= length1(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220from = \y0.2 + 2y0 a!6220!6220from# = \y0.2y0 a!6220!6220length = \y0.0 a!6220!6220length1 = \y0.0 cons = \y0y1.2 + y0 from = \y0.2 + 2y0 length = \y0.0 length1 = \y0.0 mark = \y0.y0 mark# = \y0.2y0 s = \y0.2y0 Using this interpretation, the requirements translate to: [[a!6220!6220from#(_x0)]] = 2x0 >= 2x0 = [[mark#(_x0)]] [[mark#(from(_x0))]] = 4 + 4x0 > 2x0 = [[a!6220!6220from#(mark(_x0))]] [[mark#(from(_x0))]] = 4 + 4x0 > 2x0 = [[mark#(_x0)]] [[mark#(cons(_x0, _x1))]] = 4 + 2x0 > 2x0 = [[mark#(_x0)]] [[mark#(s(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] [[a!6220!6220from(_x0)]] = 2 + 2x0 >= 2 + x0 = [[cons(mark(_x0), from(s(_x0)))]] [[a!6220!6220length(cons(_x0, _x1))]] = 0 >= 0 = [[s(a!6220!6220length1(_x1))]] [[a!6220!6220length1(_x0)]] = 0 >= 0 = [[a!6220!6220length(_x0)]] [[mark(from(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220from(mark(_x0))]] [[mark(length(_x0))]] = 0 >= 0 = [[a!6220!6220length(_x0)]] [[mark(length1(_x0))]] = 0 >= 0 = [[a!6220!6220length1(_x0)]] [[mark(cons(_x0, _x1))]] = 2 + x0 >= 2 + x0 = [[cons(mark(_x0), _x1)]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220from(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[from(_x0)]] [[a!6220!6220length(_x0)]] = 0 >= 0 = [[length(_x0)]] [[a!6220!6220length1(_x0)]] = 0 >= 0 = [[length1(_x0)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 consists of: a!6220!6220from#(X) =#> mark#(X) mark#(s(X)) =#> mark#(X) Thus, the original system is terminating if (P_4, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_1, 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 * 1 : 1 This graph has the following strongly connected components: P_5: mark#(s(X)) =#> mark#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_4, R_1, m, f) by (P_5, R_1, m, f). Thus, the original system is terminating if (P_5, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_1, minimal, formative). We apply the subterm criterion with the following projection function: nu(mark#) = 1 Thus, we can orient the dependency pairs as follows: nu(mark#(s(X))) = s(X) |> X = nu(mark#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_1, minimal, f) by ({}, R_1, 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.