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TRS Stand 20472 pair #381710128
details
property
value
status
complete
benchmark
#3.18.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n050.star.cs.uiowa.edu
space
Applicative_first_order_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.33498692513 seconds
cpu usage
5.841185649
max memory
3.19995904E8
stage attributes
key
value
output-size
17400
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 4 ms] (2) QDP (3) DependencyGraphProof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) QDPSizeChangeProof [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) QDPApplicativeOrderProof [EQUIVALENT, 28 ms] (21) QDP (22) PisEmptyProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) QDPSizeChangeProof [EQUIVALENT, 0 ms] (26) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(double, 0) -> 0 app(double, app(s, x)) -> app(s, app(s, app(double, x))) app(app(plus, 0), y) -> y app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y)) app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y)) app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, 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) 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(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y) APP(app(minus, app(s, x)), app(s, y)) -> APP(minus, x) APP(double, app(s, x)) -> APP(s, app(s, app(double, x))) APP(double, app(s, x)) -> APP(s, app(double, x)) APP(double, app(s, x)) -> APP(double, x) APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y)) APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y) APP(app(plus, app(s, x)), y) -> APP(plus, x) APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y)) APP(app(plus, app(s, x)), y) -> APP(s, y) APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, app(app(minus, x), y)), app(double, y))) APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y)) APP(app(plus, app(s, x)), y) -> APP(plus, app(app(minus, x), y)) APP(app(plus, app(s, x)), y) -> APP(app(minus, x), y) APP(app(plus, app(s, x)), y) -> APP(minus, x) APP(app(plus, app(s, x)), y) -> APP(double, y) 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)
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