TRS Standard pair #487072058

details

property	value
status	complete
benchmark	Ex1_2_Luc02c.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n189.star.cs.uiowa.edu
space	Strategy_removed_CSR_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.03756 seconds
cpu usage	4.92354
user time	4.6973
system time	0.226244
max virtual memory	1.8477524E7
max residence set size	359156.0

stage attributes

key	value
starexec-result	NO

output

NO
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 disproven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 7 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) TransformationProof [EQUIVALENT, 0 ms]
(10) QDP
(11) TransformationProof [EQUIVALENT, 0 ms]
(12) QDP
(13) NonTerminationLoopProof [COMPLETE, 0 ms]
(14) NO


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

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

   2nd(cons(X, cons(Y, Z))) -> Y
   from(X) -> cons(X, from(s(X)))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

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

   2nd(cons(X, cons(Y, Z))) -> Y
   from(X) -> cons(X, from(s(X)))

The set Q consists of the following terms:

   2nd(cons(x0, cons(x1, x2)))
   from(x0)


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

(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:

   FROM(X) -> FROM(s(X))

The TRS R consists of the following rules:

   2nd(cons(X, cons(Y, Z))) -> Y
   from(X) -> cons(X, from(s(X)))

The set Q consists of the following terms:

   2nd(cons(x0, cons(x1, x2)))
   from(x0)

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

(5) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FROM(X) -> FROM(s(X))

R is empty.
The set Q consists of the following terms:

   2nd(cons(x0, cons(x1, x2)))
   from(x0)

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

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