TRS Standard pair #615305255

details

property	value
status	complete
benchmark	4.25.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n046.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	2.4068069458 seconds
cpu usage	4.588145734
max memory	2.72273408E8

stage attributes

key	value
output-size	2178
starexec-result	"YES"

output

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


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 71 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 2 ms]
(4) QTRS
(5) RisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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

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

   rev(a) -> a
   rev(b) -> b
   rev(++(x, y)) -> ++(rev(y), rev(x))
   rev(++(x, x)) -> rev(x)

Q is empty.

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

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
rev/1)YES(
a/0)
b/0)
++/2(YES,YES)

Quasi precedence:
a > ++_2
b > ++_2


Status:
a: multiset status
b: multiset status
++_2: multiset status

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   rev(++(x, x)) -> rev(x)




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

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

   rev(a) -> a
   rev(b) -> b
   rev(++(x, y)) -> ++(rev(y), rev(x))

Q is empty.

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

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:rev_1 > ++_2 > b > a

and weight map:

   a=1
   b=1
   rev_1=0
   ++_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   rev(a) -> a
   rev(b) -> b
   rev(++(x, y)) -> ++(rev(y), rev(x))




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

(4)

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