TRS Standard pair #487067083

details

property	value
status	complete
benchmark	pair3hard.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.98169 seconds
cpu usage	8.02018
user time	7.66805
system time	0.352127
max virtual memory	1.856554E7
max residence set size	558136.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 0 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 15 ms]
(4) QDP
(5) TransformationProof [EQUIVALENT, 0 ms]
(6) QDP
(7) SemLabProof [SOUND, 125 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) AND
    (11) QDP
        (12) QDPOrderProof [EQUIVALENT, 3 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 24 ms]
        (18) QDP
        (19) PisEmptyProof [EQUIVALENT, 0 ms]
        (20) YES


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

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

   p(a(x0), p(x1, p(x2, x3))) -> p(x1, p(x0, p(a(x3), x3)))

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:

   P(a(x0), p(x1, p(x2, x3))) -> P(x1, p(x0, p(a(x3), x3)))
   P(a(x0), p(x1, p(x2, x3))) -> P(x0, p(a(x3), x3))
   P(a(x0), p(x1, p(x2, x3))) -> P(a(x3), x3)

The TRS R consists of the following rules:

   p(a(x0), p(x1, p(x2, x3))) -> p(x1, p(x0, p(a(x3), x3)))

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

(3) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   P(a(x0), p(x1, p(x2, x3))) -> P(x0, p(a(x3), x3))
   P(a(x0), p(x1, p(x2, x3))) -> P(a(x3), x3)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
P(x1, x2)  =  x2

p(x1, x2)  =  p(x2)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   p_1=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   p(a(x0), p(x1, p(x2, x3))) -> p(x1, p(x0, p(a(x3), x3)))


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

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

   P(a(x0), p(x1, p(x2, x3))) -> P(x1, p(x0, p(a(x3), x3)))

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