TRS Stand 20472 pair #381713902

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	n090.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.05484008789 seconds
cpu usage	13.990763015
max memory	1.8296832E9

stage attributes

key	value
output-size	10583
starexec-result	NO

output

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


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NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
        (12) QDP
        (13) NonTerminationLoopProof [COMPLETE, 2589 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.
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(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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