SRS_Standard 2019-03-29 03.29 pair #432290821

details

property	value
status	complete
benchmark	size-12-alpha-3-num-360.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n049.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.97675 seconds
cpu usage	19.5986
user time	18.765
system time	0.833585
max virtual memory	4.1494916E7
max residence set size	2712468.0

stage attributes

key	value
starexec-result	YES

output

19.16/5.84	YES
19.32/5.88	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
19.32/5.88	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
19.32/5.88	
19.32/5.88	
19.32/5.88	Termination w.r.t. Q of the given QTRS could be proven:
19.32/5.88	
19.32/5.88	(0) QTRS
19.32/5.88	(1) QTRS Reverse [EQUIVALENT, 0 ms]
19.32/5.88	(2) QTRS
19.32/5.88	(3) DependencyPairsProof [EQUIVALENT, 9 ms]
19.32/5.88	(4) QDP
19.32/5.88	(5) MRRProof [EQUIVALENT, 66 ms]
19.32/5.88	(6) QDP
19.32/5.88	(7) QDPOrderProof [EQUIVALENT, 0 ms]
19.32/5.88	(8) QDP
19.32/5.88	(9) DependencyGraphProof [EQUIVALENT, 0 ms]
19.32/5.88	(10) TRUE
19.32/5.88	
19.32/5.88	
19.32/5.88	----------------------------------------
19.32/5.88	
19.32/5.88	(0)
19.32/5.88	Obligation:
19.32/5.88	Q restricted rewrite system:
19.32/5.88	The TRS R consists of the following rules:
19.32/5.88	
19.32/5.88	   a(x1) -> b(x1)
19.32/5.88	   a(b(b(x1))) -> c(x1)
19.32/5.88	   c(c(x1)) -> a(b(c(a(x1))))
19.32/5.88	
19.32/5.88	Q is empty.
19.32/5.88	
19.32/5.88	----------------------------------------
19.32/5.88	
19.32/5.88	(1) QTRS Reverse (EQUIVALENT)
19.32/5.88	We applied the QTRS Reverse Processor [REVERSE].
19.32/5.88	----------------------------------------
19.32/5.88	
19.32/5.88	(2)
19.32/5.88	Obligation:
19.32/5.88	Q restricted rewrite system:
19.32/5.88	The TRS R consists of the following rules:
19.32/5.88	
19.32/5.88	   a(x1) -> b(x1)
19.32/5.88	   b(b(a(x1))) -> c(x1)
19.32/5.88	   c(c(x1)) -> a(c(b(a(x1))))
19.32/5.88	
19.32/5.88	Q is empty.
19.32/5.88	
19.32/5.88	----------------------------------------
19.32/5.88	
19.32/5.88	(3) DependencyPairsProof (EQUIVALENT)
19.32/5.88	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
19.32/5.88	----------------------------------------
19.32/5.88	
19.32/5.88	(4)
19.32/5.88	Obligation:
19.32/5.88	Q DP problem:
19.32/5.88	The TRS P consists of the following rules:
19.32/5.88	
19.32/5.88	   A(x1) -> B(x1)
19.32/5.88	   B(b(a(x1))) -> C(x1)
19.32/5.88	   C(c(x1)) -> A(c(b(a(x1))))
19.32/5.88	   C(c(x1)) -> C(b(a(x1)))
19.32/5.88	   C(c(x1)) -> B(a(x1))
19.32/5.88	   C(c(x1)) -> A(x1)
19.32/5.88	
19.32/5.88	The TRS R consists of the following rules:
19.32/5.88	
19.32/5.88	   a(x1) -> b(x1)
19.32/5.88	   b(b(a(x1))) -> c(x1)
19.32/5.88	   c(c(x1)) -> a(c(b(a(x1))))
19.32/5.88	
19.32/5.88	Q is empty.
19.32/5.88	We have to consider all minimal (P,Q,R)-chains.
19.32/5.88	----------------------------------------
19.32/5.88	
19.32/5.88	(5) MRRProof (EQUIVALENT)
19.32/5.88	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.
19.32/5.88	
19.32/5.88	Strictly oriented dependency pairs:
19.32/5.88	
19.32/5.88	   C(c(x1)) -> C(b(a(x1)))
19.32/5.88	   C(c(x1)) -> B(a(x1))
19.32/5.88	   C(c(x1)) -> A(x1)
19.32/5.88	
19.32/5.88	
19.32/5.88	Used ordering: Polynomial interpretation [POLO]:
19.32/5.88	
19.32/5.88	   POL(A(x_1)) = x_1
19.32/5.88	   POL(B(x_1)) = x_1
19.32/5.88	   POL(C(x_1)) = 2 + x_1
19.32/5.88	   POL(a(x_1)) = 1 + x_1
19.32/5.88	   POL(b(x_1)) = 1 + x_1
19.32/5.88	   POL(c(x_1)) = 3 + x_1
19.32/5.88	
19.32/5.88	
19.32/5.88	----------------------------------------
19.32/5.88

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