TRS_Relative 2019-03-21 03.54 pair #429988725

details

property	value
status	complete
benchmark	rt1-3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n105.star.cs.uiowa.edu
space	Relative_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.65246 seconds
cpu usage	3.28029
user time	3.12978
system time	0.150501
max virtual memory	1.8409688E7
max residence set size	223456.0

stage attributes

key	value
starexec-result	YES

output

3.23/1.63	YES
3.23/1.63	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.23/1.63	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.23/1.63	
3.23/1.63	
3.23/1.63	Termination of the given RelTRS could be proven:
3.23/1.63	
3.23/1.63	(0) RelTRS
3.23/1.63	(1) RelTRStoQDPProof [SOUND, 0 ms]
3.23/1.63	(2) QDP
3.23/1.63	(3) MRRProof [EQUIVALENT, 52 ms]
3.23/1.63	(4) QDP
3.23/1.63	(5) PisEmptyProof [EQUIVALENT, 0 ms]
3.23/1.63	(6) YES
3.23/1.63	
3.23/1.63	
3.23/1.63	----------------------------------------
3.23/1.63	
3.23/1.63	(0)
3.23/1.63	Obligation:
3.23/1.63	Relative term rewrite system:
3.23/1.63	The relative TRS consists of the following R rules:
3.23/1.63	
3.23/1.63	   t(f(x), g(y), f(z)) -> t(z, g(x), g(y))
3.23/1.63	   t(g(x), g(y), f(z)) -> t(f(y), f(z), x)
3.23/1.63	
3.23/1.63	The relative TRS consists of the following S rules:
3.23/1.63	
3.23/1.63	   f(g(x)) -> g(f(x))
3.23/1.63	   g(f(x)) -> f(g(x))
3.23/1.63	   f(f(x)) -> g(g(x))
3.23/1.63	   g(g(x)) -> f(f(x))
3.23/1.63	
3.23/1.63	
3.23/1.63	----------------------------------------
3.23/1.63	
3.23/1.63	(1) RelTRStoQDPProof (SOUND)
3.23/1.63	The relative termination problem is root-restricted. We can therefore treat it as a dependency pair problem.
3.23/1.63	----------------------------------------
3.23/1.63	
3.23/1.63	(2)
3.23/1.63	Obligation:
3.23/1.63	Q DP problem:
3.23/1.63	The TRS P consists of the following rules:
3.23/1.63	
3.23/1.63	   t(f(x), g(y), f(z)) -> t(z, g(x), g(y))
3.23/1.64	   t(g(x), g(y), f(z)) -> t(f(y), f(z), x)
3.23/1.64	
3.23/1.64	The TRS R consists of the following rules:
3.23/1.64	
3.23/1.64	   f(g(x)) -> g(f(x))
3.23/1.64	   g(f(x)) -> f(g(x))
3.23/1.64	   f(f(x)) -> g(g(x))
3.23/1.64	   g(g(x)) -> f(f(x))
3.23/1.64	
3.23/1.64	Q is empty.
3.23/1.64	We have to consider all (P,Q,R)-chains.
3.23/1.64	----------------------------------------
3.23/1.64	
3.23/1.64	(3) MRRProof (EQUIVALENT)
3.23/1.64	By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
3.23/1.64	
3.23/1.64	Strictly oriented dependency pairs:
3.23/1.64	
3.23/1.64	   t(f(x), g(y), f(z)) -> t(z, g(x), g(y))
3.23/1.64	   t(g(x), g(y), f(z)) -> t(f(y), f(z), x)
3.23/1.64	
3.23/1.64	
3.23/1.64	Used ordering: Polynomial interpretation [POLO]:
3.23/1.64	
3.23/1.64	   POL(f(x_1)) = 2 + 2*x_1
3.23/1.64	   POL(g(x_1)) = 2 + 2*x_1
3.23/1.64	   POL(t(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3
3.23/1.64	
3.23/1.64	
3.23/1.64	----------------------------------------
3.23/1.64	
3.23/1.64	(4)
3.23/1.64	Obligation:
3.23/1.64	Q DP problem:
3.23/1.64	P is empty.
3.23/1.64	The TRS R consists of the following rules:
3.23/1.64	
3.23/1.64	   f(g(x)) -> g(f(x))
3.23/1.64	   g(f(x)) -> f(g(x))
3.23/1.64	   f(f(x)) -> g(g(x))
3.23/1.64	   g(g(x)) -> f(f(x))
3.23/1.64	
3.23/1.64	Q is empty.
3.23/1.64	We have to consider all (P,Q,R)-chains.
3.23/1.64	----------------------------------------
3.23/1.64	
3.23/1.64	(5) PisEmptyProof (EQUIVALENT)
3.23/1.64	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.23/1.64	----------------------------------------
3.23/1.64	
3.23/1.64	(6)
3.23/1.64	YES
3.23/1.65	EOF

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