TRS Standard pair #487069258

details

property	value
status	complete
benchmark	revlist.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n187.star.cs.uiowa.edu
space	Rubio_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.82301 seconds
cpu usage	4.16669
user time	3.96175
system time	0.204944
max virtual memory	1.8544088E7
max residence set size	263480.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 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) QDPOrderProof [EQUIVALENT, 50 ms]
        (16) QDP
        (17) DependencyGraphProof [EQUIVALENT, 0 ms]
        (18) TRUE


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

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

   rev1(0, nil) -> 0
   rev1(s(X), nil) -> s(X)
   rev1(X, cons(Y, L)) -> rev1(Y, L)
   rev(nil) -> nil
   rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
   rev2(X, nil) -> nil
   rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

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:

   rev1(0, nil) -> 0
   rev1(s(X), nil) -> s(X)
   rev1(X, cons(Y, L)) -> rev1(Y, L)
   rev(nil) -> nil
   rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
   rev2(X, nil) -> nil
   rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

   rev1(0, nil)
   rev1(s(x0), nil)
   rev1(x0, cons(x1, x2))
   rev(nil)
   rev(cons(x0, x1))
   rev2(x0, nil)
   rev2(x0, cons(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:

   REV1(X, cons(Y, L)) -> REV1(Y, L)
   REV(cons(X, L)) -> REV1(X, L)
   REV(cons(X, L)) -> REV2(X, L)
   REV2(X, cons(Y, L)) -> REV(cons(X, rev(rev2(Y, L))))
   REV2(X, cons(Y, L)) -> REV(rev2(Y, L))
   REV2(X, cons(Y, L)) -> REV2(Y, L)

The TRS R consists of the following rules:

   rev1(0, nil) -> 0
   rev1(s(X), nil) -> s(X)
   rev1(X, cons(Y, L)) -> rev1(Y, L)
   rev(nil) -> nil
   rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))

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