details

property	value
status	complete
benchmark	023.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n077.star.cs.uiowa.edu
space	AotoYamada_05

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0488431453705 seconds
cpu usage	0.037635857
max memory	1839104.0

stage attributes

key	value
output-size	2274
starexec-result	YES

output

/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 app : [o * o] --> o id : [] --> o plus : [] --> o s : [] --> o app(id, X) => X app(plus, 0) => id app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(id, X) >? X app(plus, 0) >? id app(app(plus, app(s, X)), Y) >? app(s, app(app(plus, X), Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 app = \y0y1.1 + y0 + y1 id = 0 plus = 0 s = 0 Using this interpretation, the requirements translate to: [[app(id, _x0)]] = 1 + x0 > x0 = [[_x0]] [[app(plus, 0)]] = 4 > 0 = [[id]] [[app(app(plus, app(s, _x0)), _x1)]] = 3 + x0 + x1 >= 3 + x0 + x1 = [[app(s, app(app(plus, _x0), _x1))]] We can thus remove the following rules: app(id, X) => X app(plus, 0) => id We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(app(plus, app(s, X)), Y) >? app(s, app(app(plus, X), Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: app = \y0y1.1 + y1 + 2y0 plus = 0 s = 0 Using this interpretation, the requirements translate to: [[app(app(plus, app(s, _x0)), _x1)]] = 5 + x1 + 2x0 > 4 + x1 + 2x0 = [[app(s, app(app(plus, _x0), _x1))]] We can thus remove the following rules: app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) 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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