SRS Standard pair #516975556

details

property	value
status	complete
benchmark	secr10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	5.40076303482 seconds
cpu usage	18.126652056
max memory	1.320947712E9

stage attributes

key	value
output-size	5986
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 22 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 41 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 186 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(a(x1)) -> c(b(a(b(a(x1)))))
   b(a(b(x1))) -> b(x1)
   a(a(a(x1))) -> c(c(a(x1)))
   c(c(x1)) -> a(b(c(b(a(x1)))))
   a(c(a(x1))) -> c(c(a(x1)))
   c(a(c(x1))) -> a(a(c(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(a(x1)) -> C(b(a(b(a(x1)))))
   A(a(x1)) -> B(a(b(a(x1))))
   A(a(x1)) -> A(b(a(x1)))
   A(a(x1)) -> B(a(x1))
   A(a(a(x1))) -> C(c(a(x1)))
   A(a(a(x1))) -> C(a(x1))
   C(c(x1)) -> A(b(c(b(a(x1)))))
   C(c(x1)) -> B(c(b(a(x1))))
   C(c(x1)) -> C(b(a(x1)))
   C(c(x1)) -> B(a(x1))
   C(c(x1)) -> A(x1)
   A(c(a(x1))) -> C(c(a(x1)))
   C(a(c(x1))) -> A(a(c(x1)))

The TRS R consists of the following rules:

   a(a(x1)) -> c(b(a(b(a(x1)))))
   b(a(b(x1))) -> b(x1)
   a(a(a(x1))) -> c(c(a(x1)))
   c(c(x1)) -> a(b(c(b(a(x1)))))
   a(c(a(x1))) -> c(c(a(x1)))
   c(a(c(x1))) -> a(a(c(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 4 less nodes.
----------------------------------------

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

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

The TRS R consists of the following rules:

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