/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!6220first : [o * o] --> o a!6220!6220from : [o] --> o cons : [o * o] --> o first : [o * o] --> o from : [o] --> o mark : [o] --> o nil : [] --> o s : [o] --> o a!6220!6220first(0, X) => nil a!6220!6220first(s(X), cons(Y, Z)) => cons(mark(Y), first(X, Z)) a!6220!6220from(X) => cons(mark(X), from(s(X))) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(from(X)) => a!6220!6220from(mark(X)) mark(0) => 0 mark(nil) => nil mark(s(X)) => s(mark(X)) mark(cons(X, Y)) => cons(mark(X), Y) a!6220!6220first(X, Y) => first(X, Y) a!6220!6220from(X) => from(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!6220first#(s(X), cons(Y, Z)) =#> mark#(Y) 1] a!6220!6220from#(X) =#> mark#(X) 2] mark#(first(X, Y)) =#> a!6220!6220first#(mark(X), mark(Y)) 3] mark#(first(X, Y)) =#> mark#(X) 4] mark#(first(X, Y)) =#> mark#(Y) 5] mark#(from(X)) =#> a!6220!6220from#(mark(X)) 6] mark#(from(X)) =#> mark#(X) 7] mark#(s(X)) =#> mark#(X) 8] mark#(cons(X, Y)) =#> mark#(X) Rules R_0: a!6220!6220first(0, X) => nil a!6220!6220first(s(X), cons(Y, Z)) => cons(mark(Y), first(X, Z)) a!6220!6220from(X) => cons(mark(X), from(s(X))) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(from(X)) => a!6220!6220from(mark(X)) mark(0) => 0 mark(nil) => nil mark(s(X)) => s(mark(X)) mark(cons(X, Y)) => cons(mark(X), Y) a!6220!6220first(X, Y) => first(X, Y) a!6220!6220from(X) => from(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). The formative rules of (P_0, R_0) are R_1 ::= a!6220!6220first(s(X), cons(Y, Z)) => cons(mark(Y), first(X, Z)) a!6220!6220from(X) => cons(mark(X), from(s(X))) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(from(X)) => a!6220!6220from(mark(X)) mark(s(X)) => s(mark(X)) mark(cons(X, Y)) => cons(mark(X), Y) a!6220!6220first(X, Y) => first(X, Y) a!6220!6220from(X) => from(X) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative). Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_0, 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!6220first#(s(X), cons(Y, Z)) >? mark#(Y) a!6220!6220from#(X) >? mark#(X) mark#(first(X, Y)) >? a!6220!6220first#(mark(X), mark(Y)) mark#(first(X, Y)) >? mark#(X) mark#(first(X, Y)) >? mark#(Y) mark#(from(X)) >? a!6220!6220from#(mark(X)) mark#(from(X)) >? mark#(X) mark#(s(X)) >? mark#(X) mark#(cons(X, Y)) >? mark#(X) a!6220!6220first(s(X), cons(Y, Z)) >= cons(mark(Y), first(X, Z)) a!6220!6220from(X) >= cons(mark(X), from(s(X))) mark(first(X, Y)) >= a!6220!6220first(mark(X), mark(Y)) mark(from(X)) >= a!6220!6220from(mark(X)) mark(s(X)) >= s(mark(X)) mark(cons(X, Y)) >= cons(mark(X), Y) a!6220!6220first(X, Y) >= first(X, Y) a!6220!6220from(X) >= from(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220first = \y0y1.2 + 2y0 + 2y1 a!6220!6220first# = \y0y1.1 + y1 a!6220!6220from = \y0.1 + 2y0 a!6220!6220from# = \y0.1 + 2y0 cons = \y0y1.y0 first = \y0y1.2 + 2y0 + 2y1 from = \y0.1 + 2y0 mark = \y0.y0 mark# = \y0.y0 s = \y0.2y0 Using this interpretation, the requirements translate to: [[a!6220!6220first#(s(_x0), cons(_x1, _x2))]] = 1 + x1 > x1 = [[mark#(_x1)]] [[a!6220!6220from#(_x0)]] = 1 + 2x0 > x0 = [[mark#(_x0)]] [[mark#(first(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + x1 = [[a!6220!6220first#(mark(_x0), mark(_x1))]] [[mark#(first(_x0, _x1))]] = 2 + 2x0 + 2x1 > x0 = [[mark#(_x0)]] [[mark#(first(_x0, _x1))]] = 2 + 2x0 + 2x1 > x1 = [[mark#(_x1)]] [[mark#(from(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[a!6220!6220from#(mark(_x0))]] [[mark#(from(_x0))]] = 1 + 2x0 > x0 = [[mark#(_x0)]] [[mark#(s(_x0))]] = 2x0 >= x0 = [[mark#(_x0)]] [[mark#(cons(_x0, _x1))]] = x0 >= x0 = [[mark#(_x0)]] [[a!6220!6220first(s(_x0), cons(_x1, _x2))]] = 2 + 2x1 + 4x0 >= x1 = [[cons(mark(_x1), first(_x0, _x2))]] [[a!6220!6220from(_x0)]] = 1 + 2x0 >= x0 = [[cons(mark(_x0), from(s(_x0)))]] [[mark(first(_x0, _x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[a!6220!6220first(mark(_x0), mark(_x1))]] [[mark(from(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[a!6220!6220from(mark(_x0))]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[mark(cons(_x0, _x1))]] = x0 >= x0 = [[cons(mark(_x0), _x1)]] [[a!6220!6220first(_x0, _x1)]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[first(_x0, _x1)]] [[a!6220!6220from(_x0)]] = 1 + 2x0 >= 1 + 2x0 = [[from(_x0)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 consists of: mark#(from(X)) =#> a!6220!6220from#(mark(X)) mark#(s(X)) =#> mark#(X) mark#(cons(X, Y)) =#> mark#(X) 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : 0, 1, 2 * 2 : 0, 1, 2 This graph has the following strongly connected components: P_2: mark#(s(X)) =#> mark#(X) mark#(cons(X, Y)) =#> mark#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_1, R_1, m, f) by (P_2, R_1, m, f). Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, 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)) nu(mark#(cons(X, Y))) = cons(X, Y) |> X = nu(mark#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, 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.