details

property	value
status	complete
benchmark	jwcime2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n010.star.cs.uiowa.edu
space	Waldmann_06

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.462683916092 seconds
cpu usage	0.4602106
max memory	1.163264E7

stage attributes

key	value
output-size	3344
starexec-result	YES

output

/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. a : [] --> o f : [o * o] --> o h : [o] --> o f(X, f(Y, a)) => f(f(f(f(a, a), Y), h(a)), 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] f#(X, f(Y, a)) =#> f#(f(f(f(a, a), Y), h(a)), X) 1] f#(X, f(Y, a)) =#> f#(f(f(a, a), Y), h(a)) 2] f#(X, f(Y, a)) =#> f#(f(a, a), Y) 3] f#(X, f(Y, a)) =#> f#(a, a) Rules R_0: f(X, f(Y, a)) => f(f(f(f(a, a), Y), h(a)), 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 : 0, 1, 2, 3 * 1 : * 2 : 0, 1, 2, 3 * 3 : This graph has the following strongly connected components: P_1: f#(X, f(Y, a)) =#> f#(f(f(f(a, a), Y), h(a)), X) f#(X, f(Y, a)) =#> f#(f(a, a), 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). 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 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: f#(X, f(Y, a)) >? f#(f(f(f(a, a), Y), h(a)), X) f#(X, f(Y, a)) >? f#(f(a, a), Y) f(X, f(Y, a)) >= f(f(f(f(a, a), Y), h(a)), X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a = 1 f = \y0y1.2y1 + 3y0y1 f# = \y0y1.y0 + 3y1 + y0y1 h = \y0.0 Using this interpretation, the requirements translate to: [[f#(_x0, f(_x1, a))]] = 6 + 3x0 + 3x0x1 + 9x1 > 3x0 = [[f#(f(f(f(a, a), _x1), h(a)), _x0)]] [[f#(_x0, f(_x1, a))]] = 6 + 3x0 + 3x0x1 + 9x1 > 5 + 8x1 = [[f#(f(a, a), _x1)]] [[f(_x0, f(_x1, a))]] = 4 + 6x0 + 6x1 + 9x0x1 >= 2x0 = [[f(f(f(f(a, a), _x1), h(a)), _x0)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0). 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. popout

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