TRS_Equational 2019-03-21 05.09 pair #429997297

details

property	value
status	complete
benchmark	boolean_rings.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	Mixed_AC

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.67086 seconds
cpu usage	3.22255
user time	3.0919
system time	0.130651
max virtual memory	1.8275288E7
max residence set size	218792.0

stage attributes

key	value
starexec-result	YES

output

2.85/1.61	YES
3.18/1.62	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
3.18/1.62	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.18/1.62	
3.18/1.62	
3.18/1.62	Termination of the given ETRS could be proven:
3.18/1.62	
3.18/1.62	(0) ETRS
3.18/1.62	(1) EDirectTerminationProof [EQUIVALENT, 0 ms]
3.18/1.62	(2) YES
3.18/1.62	
3.18/1.62	
3.18/1.62	----------------------------------------
3.18/1.62	
3.18/1.62	(0)
3.18/1.62	Obligation:
3.18/1.62	Equational rewrite system:
3.18/1.62	The TRS R consists of the following rules:
3.18/1.62	
3.18/1.62	   xor(F, x) -> x
3.18/1.62	   xor(neg(x), x) -> F
3.18/1.62	   and(T, x) -> x
3.18/1.62	   and(F, x) -> F
3.18/1.62	   and(x, x) -> x
3.18/1.62	   and(xor(x, y), z) -> xor(and(x, z), and(y, z))
3.18/1.62	   xor(x, x) -> F
3.18/1.62	   impl(x, y) -> xor(and(x, y), xor(T, x))
3.18/1.62	   or(x, y) -> xor(and(x, y), xor(x, y))
3.18/1.62	   equiv(x, y) -> xor(xor(T, y), x)
3.18/1.62	   neg(x) -> xor(T, x)
3.18/1.62	
3.18/1.62	The set E consists of the following equations:
3.18/1.62	
3.18/1.62	   and(x, y) == and(y, x)
3.18/1.62	   or(x, y) == or(y, x)
3.18/1.62	   xor(x, y) == xor(y, x)
3.18/1.62	   and(and(x, y), z) == and(x, and(y, z))
3.18/1.62	   or(or(x, y), z) == or(x, or(y, z))
3.18/1.62	   xor(xor(x, y), z) == xor(x, xor(y, z))
3.18/1.62	
3.18/1.62	
3.18/1.62	----------------------------------------
3.18/1.62	
3.18/1.62	(1) EDirectTerminationProof (EQUIVALENT)
3.18/1.62	We use [DA_FALKE] with the following order to prove termination.
3.18/1.62	
3.18/1.62	Precedence:
3.18/1.62	neg_1 > xor_2 > F
3.18/1.62	neg_1 > T
3.18/1.62	impl_2 > and_2 > xor_2 > F
3.18/1.62	impl_2 > T
3.18/1.62	or_2 > and_2 > xor_2 > F
3.18/1.62	equiv_2 > xor_2 > F
3.18/1.62	equiv_2 > T
3.18/1.62	
3.18/1.62	
3.18/1.62	Status:
3.18/1.62	and_2: flat status
3.18/1.62	or_2: flat status
3.18/1.62	xor_2: flat status
3.18/1.62	
3.18/1.62	
3.18/1.62	
3.18/1.62	----------------------------------------
3.18/1.62	
3.18/1.62	(2)
3.18/1.62	YES
3.18/1.67	EOF

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