TRS Standard pair #487071753

details

property	value
status	complete
benchmark	Ex1_GM99_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n183.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	7.20794 seconds
cpu usage	14.7583
user time	14.1597
system time	0.598586
max virtual memory	1.8559572E7
max residence set size	1700344.0

stage attributes

key	value
starexec-result	NO

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 6 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [EQUIVALENT, 6 ms]
        (12) QDP
        (13) NonTerminationLoopProof [COMPLETE, 2451 ms]
        (14) NO


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

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

Q is empty.

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(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:

   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))
   ACTIVE(f(a, b, X)) -> F(X, X, X)
   ACTIVE(c) -> MARK(a)
   ACTIVE(c) -> MARK(b)
   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, X2, mark(X3)))
   MARK(f(X1, X2, X3)) -> F(X1, X2, mark(X3))
   MARK(f(X1, X2, X3)) -> MARK(X3)
   MARK(a) -> ACTIVE(a)
   MARK(b) -> ACTIVE(b)
   MARK(c) -> ACTIVE(c)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, mark(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
   F(active(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, active(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, active(X3)) -> F(X1, X2, X3)

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 7 less nodes.
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(4)

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