SRS Standard pair #487091146

details

property	value
status	complete
benchmark	z103.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.78234 seconds
cpu usage	8.18007
user time	7.79271
system time	0.387363
max virtual memory	1.9941468E7
max residence set size	973808.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 4 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 13 ms]
(6) QDP
(7) MRRProof [EQUIVALENT, 0 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) TRUE


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

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

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

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(d(b(x1))) -> A(c(x1))
   A(d(x1)) -> B(x1)
   A(x1) -> B(b(b(x1)))
   A(x1) -> B(b(x1))
   A(x1) -> B(x1)

The TRS R consists of the following rules:

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

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

(5) 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:

   A(d(x1)) -> B(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1

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