SRS Stand 10685 pair #381718358

details

property	value
status	complete
benchmark	beans4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n020.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.85273885727 seconds
cpu usage	19.99567436
max memory	2.033995776E9

stage attributes

key	value
output-size	3205
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 1 ms]
(2) QDP
(3) MRRProof [EQUIVALENT, 56 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) TRUE


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   b(c(a(x1))) -> a(b(c(x1)))
   c(b(x1)) -> b(a(x1))
   a(c(b(x1))) -> c(b(a(x1)))

Q is empty.

----------------------------------------

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(a(a(x1))) -> A(b(c(x1)))
   B(a(a(x1))) -> B(c(x1))
   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> A(c(x1))
   C(a(x1)) -> C(x1)
   B(c(a(x1))) -> A(b(c(x1)))
   B(c(a(x1))) -> B(c(x1))
   B(c(a(x1))) -> C(x1)
   C(b(x1)) -> B(a(x1))
   C(b(x1)) -> A(x1)
   A(c(b(x1))) -> C(b(a(x1)))
   A(c(b(x1))) -> B(a(x1))
   A(c(b(x1))) -> A(x1)

The TRS R consists of the following rules:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   b(c(a(x1))) -> a(b(c(x1)))
   c(b(x1)) -> b(a(x1))
   a(c(b(x1))) -> c(b(a(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) MRRProof (EQUIVALENT)
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.

Strictly oriented dependency pairs:

   B(a(a(x1))) -> A(b(c(x1)))
   B(a(a(x1))) -> B(c(x1))
   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> A(c(x1))
   C(a(x1)) -> C(x1)
   B(c(a(x1))) -> A(b(c(x1)))
   B(c(a(x1))) -> B(c(x1))
   B(c(a(x1))) -> C(x1)
   C(b(x1)) -> A(x1)
   A(c(b(x1))) -> B(a(x1))
   A(c(b(x1))) -> A(x1)

Strictly oriented rules of the TRS R:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   a(c(b(x1))) -> c(b(a(x1)))

Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2*x_1

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