SRS Standard pair #487083052

details

property	value
status	complete
benchmark	10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n146.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.15251 seconds
cpu usage	13.4255
user time	12.8017
system time	0.623755
max virtual memory	7.8899764E7
max residence set size	1725044.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 24 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 97 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 52 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)))
   d(x1) -> x1
   a(x1) -> x1
   b(d(b(x1))) -> a(d(x1))
   b(c(x1)) -> c(d(d(x1)))
   a(c(x1)) -> b(b(c(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)) -> D(b(x1))
   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(d(x1))
   B(d(b(x1))) -> D(x1)
   B(c(x1)) -> D(d(x1))
   B(c(x1)) -> D(x1)
   A(c(x1)) -> B(b(c(d(x1))))
   A(c(x1)) -> B(c(d(x1)))
   A(c(x1)) -> D(x1)

The TRS R consists of the following rules:

   a(d(x1)) -> d(b(x1))
   a(x1) -> b(b(b(x1)))
   d(x1) -> x1
   a(x1) -> x1
   b(d(b(x1))) -> a(d(x1))
   b(c(x1)) -> c(d(d(x1)))
   a(c(x1)) -> b(b(c(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 6 less nodes.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

popout

output may be truncated. 'popout' for the full output.

job log

popout

actions all output return to SRS Standard