TRS_Innermost 2019-03-28 22.12 pair #432271440

details

property	value
status	complete
benchmark	#4.5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n128.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.97293 seconds
cpu usage	4.5847
user time	4.36598
system time	0.218717
max virtual memory	1.8948764E7
max residence set size	375240.0

stage attributes

key	value
starexec-result	YES

output

4.27/1.93	YES
4.51/1.94	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
4.51/1.94	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.51/1.94	
4.51/1.94	
4.51/1.94	Termination w.r.t. Q of the given QTRS could be proven:
4.51/1.94	
4.51/1.94	(0) QTRS
4.51/1.94	(1) QTRSRRRProof [EQUIVALENT, 43 ms]
4.51/1.94	(2) QTRS
4.51/1.94	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
4.51/1.94	(4) QDP
4.51/1.94	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
4.51/1.94	(6) TRUE
4.51/1.94	
4.51/1.94	
4.51/1.94	----------------------------------------
4.51/1.94	
4.51/1.94	(0)
4.51/1.94	Obligation:
4.51/1.94	Q restricted rewrite system:
4.51/1.94	The TRS R consists of the following rules:
4.51/1.94	
4.51/1.94	   f(0) -> f(0)
4.51/1.94	   0 -> 1
4.51/1.94	
4.51/1.94	The set Q consists of the following terms:
4.51/1.94	
4.51/1.94	   0
4.51/1.94	
4.51/1.94	
4.51/1.94	----------------------------------------
4.51/1.94	
4.51/1.94	(1) QTRSRRRProof (EQUIVALENT)
4.51/1.94	Used ordering:
4.51/1.94	Polynomial interpretation [POLO]:
4.51/1.94	
4.51/1.94	   POL(0) = 1
4.51/1.94	   POL(1) = 0
4.51/1.94	   POL(f(x_1)) = x_1
4.51/1.94	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
4.51/1.94	
4.51/1.94	   0 -> 1
4.51/1.94	
4.51/1.94	
4.51/1.94	
4.51/1.94	
4.51/1.94	----------------------------------------
4.51/1.94	
4.51/1.94	(2)
4.51/1.94	Obligation:
4.51/1.94	Q restricted rewrite system:
4.51/1.94	The TRS R consists of the following rules:
4.51/1.94	
4.51/1.94	   f(0) -> f(0)
4.51/1.94	
4.51/1.94	The set Q consists of the following terms:
4.51/1.94	
4.51/1.94	   0
4.51/1.94	
4.51/1.94	
4.51/1.94	----------------------------------------
4.51/1.94	
4.51/1.94	(3) DependencyPairsProof (EQUIVALENT)
4.51/1.94	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
4.51/1.94	----------------------------------------
4.51/1.94	
4.51/1.94	(4)
4.51/1.94	Obligation:
4.51/1.94	Q DP problem:
4.51/1.94	The TRS P consists of the following rules:
4.51/1.94	
4.51/1.94	   F(0) -> F(0)
4.51/1.94	
4.51/1.94	The TRS R consists of the following rules:
4.51/1.94	
4.51/1.94	   f(0) -> f(0)
4.51/1.94	
4.51/1.94	The set Q consists of the following terms:
4.51/1.94	
4.51/1.94	   0
4.51/1.94	
4.51/1.94	We have to consider all minimal (P,Q,R)-chains.
4.51/1.94	----------------------------------------
4.51/1.94	
4.51/1.94	(5) DependencyGraphProof (EQUIVALENT)
4.51/1.94	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
4.51/1.94	----------------------------------------
4.51/1.94	
4.51/1.94	(6)
4.51/1.94	TRUE
4.56/1.97	EOF

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