details

property	value
status	complete
benchmark	koen.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n072.star.cs.uiowa.edu
space	Rubio_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.035619020462 seconds
cpu usage	0.025263146
max memory	1527808.0

stage attributes

key	value
output-size	2057
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] --> o c : [o] --> o f : [o * o] --> o s : [o] --> o f(s(X), X) => f(X, a(X)) f(X, c(X)) => f(s(X), X) f(X, X) => c(X) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(s(X), X) >? f(X, a(X)) f(X, c(X)) >? f(s(X), X) f(X, X) >? c(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a = \y0.y0 c = \y0.3 + 3y0 f = \y0y1.3 + 2y0 + 3y1 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[f(s(_x0), _x0)]] = 5 + 7x0 > 3 + 5x0 = [[f(_x0, a(_x0))]] [[f(_x0, c(_x0))]] = 12 + 11x0 > 5 + 7x0 = [[f(s(_x0), _x0)]] [[f(_x0, _x0)]] = 3 + 5x0 >= 3 + 3x0 = [[c(_x0)]] We can thus remove the following rules: f(s(X), X) => f(X, a(X)) f(X, c(X)) => f(s(X), X) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(X, X) >? c(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: c = \y0.y0 f = \y0y1.3 + 3y0 + 3y1 Using this interpretation, the requirements translate to: [[f(_x0, _x0)]] = 3 + 6x0 > x0 = [[c(_x0)]] We can thus remove the following rules: f(X, X) => c(X) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. popout

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