TRS Inner 89993 pair #381733468

details

property	value
status	complete
benchmark	#4.8.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n052.star.cs.uiowa.edu
space	Applicative_AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.11711001396 seconds
cpu usage	7.675284851
max memory	4.57842688E8

stage attributes

key	value
output-size	17763
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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


YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 39 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) TransformationProof [EQUIVALENT, 0 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(app(plus, x), 0) -> x
   app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y))
   app(app(app(f, 0), app(s, 0)), x) -> app(app(app(f, x), app(app(plus, x), x)), x)
   app(app(g, x), y) -> x
   app(app(g, x), y) -> y
   app(app(map, fun), nil) -> nil
   app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs))
   app(app(filter, fun), nil) -> nil
   app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs)
   app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs))
   app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs)

The set Q consists of the following terms:

   app(app(plus, x0), 0)
   app(app(plus, x0), app(s, x1))
   app(app(app(f, 0), app(s, 0)), x0)
   app(app(g, 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(plus, x), app(s, y)) -> APP(s, app(app(plus, x), y))
   APP(app(plus, x), app(s, y)) -> APP(app(plus, x), y)
   APP(app(app(f, 0), app(s, 0)), x) -> APP(app(app(f, x), app(app(plus, x), x)), x)
   APP(app(app(f, 0), app(s, 0)), x) -> APP(app(f, x), app(app(plus, x), x))
   APP(app(app(f, 0), app(s, 0)), x) -> APP(f, x)
   APP(app(app(f, 0), app(s, 0)), x) -> APP(app(plus, x), x)
   APP(app(app(f, 0), app(s, 0)), x) -> APP(plus, x)
   APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs))
   APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x))
   APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x)
   APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs)
   APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
   APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x)

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