TRS Standard pair #487072453

details

property	value
status	complete
benchmark	17.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n190.star.cs.uiowa.edu
space	Applicative_first_order_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.00301 seconds
cpu usage	4.69027
user time	4.47135
system time	0.218918
max virtual memory	1.8475476E7
max residence set size	285440.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 8 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) UsableRulesReductionPairsProof [EQUIVALENT, 9 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) ATransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (21) YES


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

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

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))
   app(i, 1) -> 1
   app(i, app(i, x)) -> x
   app(i, app(app(., x), y)) -> app(app(., app(i, y)), app(i, x))
   app(app(map, f), nil) -> nil
   app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
   app(app(filter, f), nil) -> nil
   app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs)
   app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs))
   app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs)

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP(app(., app(app(., x), y)), z) -> APP(app(., x), app(app(., y), z))
   APP(app(., app(app(., x), y)), z) -> APP(app(., y), z)
   APP(app(., app(app(., x), y)), z) -> APP(., y)
   APP(i, app(app(., x), y)) -> APP(app(., app(i, y)), app(i, x))
   APP(i, app(app(., x), y)) -> APP(., app(i, y))
   APP(i, app(app(., x), y)) -> APP(i, y)
   APP(i, app(app(., x), y)) -> APP(i, x)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
   APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x))
   APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f)
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x))
   APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs))
   APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs)
   APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f)

The TRS R consists of the following rules:

   app(app(., 1), x) -> x
   app(app(., x), 1) -> x
   app(app(., app(i, x)), x) -> 1
   app(app(., x), app(i, x)) -> 1
   app(app(., app(i, y)), app(app(., y), z)) -> z
   app(app(., y), app(app(., app(i, y)), z)) -> z
   app(app(., app(app(., x), y)), z) -> app(app(., x), app(app(., y), z))

popout

output may be truncated. 'popout' for the full output.

job log

popout

actions all output return to TRS Standard