TRS_Innermost 2019-03-28 22.12 pair #432270768

details

property	value
status	complete
benchmark	innermost2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n090.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.81638 seconds
cpu usage	3.72548
user time	3.55085
system time	0.174632
max virtual memory	1.8474652E7
max residence set size	238624.0

stage attributes

key	value
starexec-result	YES

output

3.65/1.78	YES
3.65/1.79	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.65/1.79	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.65/1.79	
3.65/1.79	
3.65/1.79	Termination w.r.t. Q of the given QTRS could be proven:
3.65/1.79	
3.65/1.79	(0) QTRS
3.65/1.79	(1) QTRSRRRProof [EQUIVALENT, 69 ms]
3.65/1.79	(2) QTRS
3.65/1.79	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.65/1.79	(4) QDP
3.65/1.79	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.65/1.79	(6) TRUE
3.65/1.79	
3.65/1.79	
3.65/1.79	----------------------------------------
3.65/1.79	
3.65/1.79	(0)
3.65/1.79	Obligation:
3.65/1.79	Q restricted rewrite system:
3.65/1.79	The TRS R consists of the following rules:
3.65/1.79	
3.65/1.79	   g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x))
3.65/1.79	   f(a(x)) -> c
3.65/1.79	   a(b) -> d
3.65/1.79	
3.65/1.79	The set Q consists of the following terms:
3.65/1.79	
3.65/1.79	   g(c, g(a(x0), x1))
3.65/1.79	   f(a(x0))
3.65/1.79	   a(b)
3.65/1.79	
3.65/1.79	
3.65/1.79	----------------------------------------
3.65/1.79	
3.65/1.79	(1) QTRSRRRProof (EQUIVALENT)
3.65/1.79	Used ordering:
3.65/1.79	Polynomial interpretation [POLO]:
3.65/1.79	
3.65/1.79	   POL(a(x_1)) = 1 + x_1
3.65/1.79	   POL(b) = 0
3.65/1.79	   POL(c) = 1
3.65/1.79	   POL(d) = 0
3.65/1.79	   POL(f(x_1)) = x_1
3.65/1.79	   POL(g(x_1, x_2)) = 2*x_1 + 2*x_2
3.65/1.79	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
3.65/1.79	
3.65/1.79	   a(b) -> d
3.65/1.79	
3.65/1.79	
3.65/1.79	
3.65/1.79	
3.65/1.79	----------------------------------------
3.65/1.79	
3.65/1.79	(2)
3.65/1.79	Obligation:
3.65/1.79	Q restricted rewrite system:
3.65/1.79	The TRS R consists of the following rules:
3.65/1.79	
3.65/1.79	   g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x))
3.65/1.79	   f(a(x)) -> c
3.65/1.79	
3.65/1.79	The set Q consists of the following terms:
3.65/1.79	
3.65/1.79	   g(c, g(a(x0), x1))
3.65/1.79	   f(a(x0))
3.65/1.79	   a(b)
3.65/1.79	
3.65/1.79	
3.65/1.79	----------------------------------------
3.65/1.79	
3.65/1.79	(3) DependencyPairsProof (EQUIVALENT)
3.65/1.79	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.65/1.79	----------------------------------------
3.65/1.79	
3.65/1.79	(4)
3.65/1.79	Obligation:
3.65/1.79	Q DP problem:
3.65/1.79	The TRS P consists of the following rules:
3.65/1.79	
3.65/1.79	   G(c, g(a(x), y)) -> G(f(a(b)), g(a(y), x))
3.65/1.79	   G(c, g(a(x), y)) -> F(a(b))
3.65/1.79	   G(c, g(a(x), y)) -> G(a(y), x)
3.65/1.79	
3.65/1.79	The TRS R consists of the following rules:
3.65/1.79	
3.65/1.79	   g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x))
3.65/1.79	   f(a(x)) -> c
3.65/1.79	
3.65/1.79	The set Q consists of the following terms:
3.65/1.79	
3.65/1.79	   g(c, g(a(x0), x1))
3.65/1.79	   f(a(x0))
3.65/1.79	   a(b)
3.65/1.79	
3.65/1.79	We have to consider all minimal (P,Q,R)-chains.
3.65/1.79	----------------------------------------
3.65/1.79	
3.65/1.79	(5) DependencyGraphProof (EQUIVALENT)

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actions all output return to TRS_Innermost 2019-03-28 22.12