TRS Innermost pair #487092919

details

property	value
status	complete
benchmark	#4.26.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	Applicative_AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.19182 seconds
cpu usage	5.51502
user time	5.25712
system time	0.257902
max virtual memory	1.8411988E7
max residence set size	354908.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 18 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) ATransformationProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) ATransformationProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QReductionProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPOrderProof [EQUIVALENT, 51 ms]
        (22) QDP
        (23) DependencyGraphProof [EQUIVALENT, 0 ms]
        (24) TRUE
    (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES


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

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

   app(p, 0) -> 0
   app(p, app(s, x)) -> x
   app(app(le, 0), y) -> true
   app(app(le, app(s, x)), 0) -> false
   app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
   app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
   app(app(app(if, true), x), y) -> 0
   app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, x)), y))
   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)

The set Q consists of the following terms:

   app(p, 0)
   app(p, app(s, x0))
   app(app(le, 0), x0)
   app(app(le, app(s, x0)), 0)
   app(app(le, app(s, x0)), app(s, x1))
   app(app(minus, x0), x1)
   app(app(app(if, true), x0), x1)
   app(app(app(if, false), x0), x1)
   app(app(map, x0), nil)
   app(app(map, x0), app(app(cons, x1), x2))
   app(app(filter, x0), nil)
   app(app(filter, x0), app(app(cons, x1), x2))
   app(app(app(app(filter2, true), x0), x1), x2)
   app(app(app(app(filter2, false), x0), x1), x2)


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

(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(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)
   APP(app(le, app(s, x)), app(s, y)) -> APP(le, x)
   APP(app(minus, x), y) -> APP(app(app(if, app(app(le, x), y)), x), y)
   APP(app(minus, x), y) -> APP(app(if, app(app(le, x), y)), x)
   APP(app(minus, x), y) -> APP(if, app(app(le, x), y))
   APP(app(minus, x), y) -> APP(app(le, x), y)
   APP(app(minus, x), y) -> APP(le, x)
   APP(app(app(if, false), x), y) -> APP(s, app(app(minus, app(p, x)), y))
   APP(app(app(if, false), x), y) -> APP(app(minus, app(p, x)), y)
   APP(app(app(if, false), x), y) -> APP(minus, app(p, x))
   APP(app(app(if, false), x), y) -> APP(p, x)
   APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))

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