TRS Stand 20472 pair #381713959

details

property	value
status	complete
benchmark	enger-nonloop-swapX.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n101.star.cs.uiowa.edu
space	EEG_IJCAR_12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	11.8365600109 seconds
cpu usage	37.247676931
max memory	4.014944256E9

stage attributes

key	value
output-size	7776
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) MNOCProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) NonLoopProof [COMPLETE, 26 ms]
        (18) NO


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

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

   g(tt, x) -> g(f(x, x), s(x))
   f(0, y) -> tt
   f(s(x), y) -> f(x, s(y))

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   f(0, y) -> tt
   f(s(x), y) -> f(x, s(y))

The TRS R 2 is 
   g(tt, x) -> g(f(x, x), s(x))

The signature Sigma is {g_2}
----------------------------------------

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

   g(tt, x) -> g(f(x, x), s(x))
   f(0, y) -> tt
   f(s(x), y) -> f(x, s(y))

The set Q consists of the following terms:

   g(tt, x0)
   f(0, x0)
   f(s(x0), x1)


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

(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:

   G(tt, x) -> G(f(x, x), s(x))
   G(tt, x) -> F(x, x)
   F(s(x), y) -> F(x, s(y))

The TRS R consists of the following rules:

   g(tt, x) -> g(f(x, x), s(x))
   f(0, y) -> tt
   f(s(x), y) -> f(x, s(y))

The set Q consists of the following terms:

   g(tt, x0)

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