TRS Inner 89993 pair #381733501

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	n003.star.cs.uiowa.edu
space	Applicative_AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.1667330265 seconds
cpu usage	5.46794576
max memory	3.60194048E8

stage attributes

key	value
output-size	20181
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 23 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, 44 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)

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