TRS Stand 20472 pair #381714847

details

property	value
status	complete
benchmark	4.24.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n042.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.83505487442 seconds
cpu usage	4.373726509
max memory	2.64470528E8

stage attributes

key	value
output-size	7770
starexec-result	YES

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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YES
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 proven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 15 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) QDPOrderProof [EQUIVALENT, 20 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

   rev(nil) -> nil
   rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
   rev1(x, nil) -> x
   rev1(x, ++(y, z)) -> rev1(y, z)
   rev2(x, nil) -> nil
   rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

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

   rev(nil) -> nil
   rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
   rev1(x, nil) -> x
   rev1(x, ++(y, z)) -> rev1(y, z)
   rev2(x, nil) -> nil
   rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))

The set Q consists of the following terms:

   rev(nil)
   rev(++(x0, x1))
   rev1(x0, nil)
   rev1(x0, ++(x1, x2))
   rev2(x0, nil)
   rev2(x0, ++(x1, x2))


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

(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:

   REV(++(x, y)) -> REV1(x, y)
   REV(++(x, y)) -> REV2(x, y)
   REV1(x, ++(y, z)) -> REV1(y, z)
   REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
   REV2(x, ++(y, z)) -> REV(rev2(y, z))
   REV2(x, ++(y, z)) -> REV2(y, z)

The TRS R consists of the following rules:

   rev(nil) -> nil
   rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))

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