TRS Standard pair #487068463

details

property	value
status	complete
benchmark	z25.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	Zantema_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.95615 seconds
cpu usage	3.87692
user time	3.70042
system time	0.176494
max virtual memory	1.8475476E7
max residence set size	246952.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

   f(a, f(b, x)) -> f(b, f(a, x))
   f(b, f(c, x)) -> f(c, f(b, x))
   f(c, f(a, x)) -> f(a, f(c, x))

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:

   F(a, f(b, x)) -> F(b, f(a, x))
   F(a, f(b, x)) -> F(a, x)
   F(b, f(c, x)) -> F(c, f(b, x))
   F(b, f(c, x)) -> F(b, x)
   F(c, f(a, x)) -> F(a, f(c, x))
   F(c, f(a, x)) -> F(c, x)

The TRS R consists of the following rules:

   f(a, f(b, x)) -> f(b, f(a, x))
   f(b, f(c, x)) -> f(c, f(b, x))
   f(c, f(a, x)) -> f(a, f(c, x))

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

(3) UsableRulesReductionPairsProof (EQUIVALENT)
First, we A-transformed [FROCOS05] the QDP-Problem.
Then we obtain the following A-transformed DP problem.

The pairs P are: 

   a1(b(x)) -> b1(a(x))
   a1(b(x)) -> a1(x)
   b1(c(x)) -> c1(b(x))
   b1(c(x)) -> b1(x)
   c1(a(x)) -> a1(c(x))
   c1(a(x)) -> c1(x)

and the Q and R are:
Q restricted rewrite system:
The TRS R consists of the following rules:

   b(c(x)) -> c(b(x))
   c(a(x)) -> a(c(x))
   a(b(x)) -> b(a(x))

Q is empty.

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

   f(a, f(b, x)) -> f(b, f(a, x))
   f(b, f(c, x)) -> f(c, f(b, x))

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