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TRS Innermost pair #487523961
details
property
value
status
complete
benchmark
#4.28.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)
2.882376194 seconds
cpu usage
4.641426854
max memory
2.70823424E8
stage attributes
key
value
output-size
18059
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 43 ms] (2) QDP (3) DependencyGraphProof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) MRRProof [EQUIVALENT, 0 ms] (22) QDP (23) DependencyGraphProof [EQUIVALENT, 0 ms] (24) TRUE (25) QDP (26) UsableRulesProof [EQUIVALENT, 0 ms] (27) QDP (28) QDPSizeChangeProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(half, 0) -> 0 app(half, app(s, 0)) -> 0 app(half, app(s, app(s, x))) -> app(s, app(half, x)) app(bits, 0) -> 0 app(bits, app(s, x)) -> app(s, app(bits, app(half, app(s, 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) The set Q consists of the following terms: app(half, 0) app(half, app(s, 0)) app(half, app(s, app(s, x0))) app(bits, 0) app(bits, app(s, x0)) 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(half, app(s, app(s, x))) -> APP(s, app(half, x)) APP(half, app(s, app(s, x))) -> APP(half, x) APP(bits, app(s, x)) -> APP(s, app(bits, app(half, app(s, x)))) APP(bits, app(s, x)) -> APP(bits, app(half, app(s, x))) APP(bits, app(s, x)) -> APP(half, app(s, 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)
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