TRS Standard pair #487069218

details

property	value
status	complete
benchmark	test4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n191.star.cs.uiowa.edu
space	Rubio_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.72169 seconds
cpu usage	3.91811
user time	3.73652
system time	0.181592
max virtual memory	1.8477524E7
max residence set size	252952.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QReductionProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 0 ms]
(10) QDP
(11) DependencyGraphProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(14) YES


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

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

   f(a, a) -> f(a, b)
   f(a, b) -> f(s(a), c)
   f(s(X), c) -> f(X, c)
   f(c, c) -> f(a, a)

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is 
   f(a, a) -> f(a, b)
   f(a, b) -> f(s(a), c)
   f(s(X), c) -> f(X, c)
   f(c, c) -> f(a, a)

The signature Sigma is {f_2}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(a, a) -> f(a, b)
   f(a, b) -> f(s(a), c)
   f(s(X), c) -> f(X, c)
   f(c, c) -> f(a, a)

The set Q consists of the following terms:

   f(a, a)
   f(a, b)
   f(s(x0), c)
   f(c, c)


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

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

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

   F(a, a) -> F(a, b)
   F(a, b) -> F(s(a), c)
   F(s(X), c) -> F(X, c)
   F(c, c) -> F(a, a)

The TRS R consists of the following rules:

   f(a, a) -> f(a, b)
   f(a, b) -> f(s(a), c)
   f(s(X), c) -> f(X, c)
   f(c, c) -> f(a, a)

The set Q consists of the following terms:

   f(a, a)
   f(a, b)
   f(s(x0), c)
   f(c, c)

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