TRS Standard pair #516961200

details

property	value
status	complete
benchmark	Ex24_Luc06_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n112.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	5.39487195015 seconds
cpu usage	13.625422536
max memory	7.61659392E8

stage attributes

key	value
output-size	9384
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be disproven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 7 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) MRRProof [EQUIVALENT, 30 ms]
        (12) QDP
        (13) NonTerminationLoopProof [COMPLETE, 504 ms]
        (14) NO


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

   active(f(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   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(b, X, c)) -> MARK(f(X, c, X))
   ACTIVE(f(b, X, c)) -> F(X, c, X)
   ACTIVE(c) -> MARK(b)
   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, mark(X2), X3))
   MARK(f(X1, X2, X3)) -> F(X1, mark(X2), X3)
   MARK(f(X1, X2, X3)) -> MARK(X2)
   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(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   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 5 less nodes.
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(4)

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