details

property	value
status	complete
benchmark	#3.26.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	AG01

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0503900051117 seconds
cpu usage	0.047849483
max memory	1916928.0

stage attributes

key	value
output-size	2457
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. f : [o] --> o s : [o] --> o f(X) => s(X) f(s(s(X))) => s(f(f(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#(s(s(X))) =#> f#(f(X)) 1] f#(s(s(X))) =#> f#(X) Rules R_0: f(X) => s(X) f(s(s(X))) => s(f(f(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 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#(s(s(X))) >? f#(f(X)) f#(s(s(X))) >? f#(X) f(X) >= s(X) f(s(s(X))) >= s(f(f(X))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0.2 + 2y0 f# = \y0.2y0 s = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[f#(s(s(_x0)))]] = 12 + 8x0 > 4 + 4x0 = [[f#(f(_x0))]] [[f#(s(s(_x0)))]] = 12 + 8x0 > 2x0 = [[f#(_x0)]] [[f(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[s(_x0)]] [[f(s(s(_x0)))]] = 14 + 8x0 >= 14 + 8x0 = [[s(f(f(_x0)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, 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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