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TRS Stand 20472 pair #381711612
details
property
value
status
complete
benchmark
13.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n108.star.cs.uiowa.edu
space
Applicative_first_order_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
13.6600399017 seconds
cpu usage
26.159716679
max memory
4.189712384E9
stage attributes
key
value
output-size
20684
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 63 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPApplicativeOrderProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 0 ms] (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) ATransformationProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) UsableRulesProof [EQUIVALENT, 0 ms] (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(*, x), app(app(+, y), z)) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(+, y), z)), x) -> app(app(+, app(app(*, x), y)), app(app(*, x), z)) app(app(*, app(app(*, x), y)), z) -> app(app(*, x), app(app(*, y), z)) app(app(+, app(app(+, x), y)), z) -> app(app(+, x), app(app(+, y), z)) 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(*, x), app(app(+, y), z)) -> APP(app(+, app(app(*, x), y)), app(app(*, x), z)) APP(app(*, x), app(app(+, y), z)) -> APP(+, app(app(*, x), y)) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), y) APP(app(*, x), app(app(+, y), z)) -> APP(app(*, x), z) APP(app(*, app(app(+, y), z)), x) -> APP(app(+, app(app(*, x), y)), app(app(*, x), z)) APP(app(*, app(app(+, y), z)), x) -> APP(+, app(app(*, x), y)) APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), y) APP(app(*, app(app(+, y), z)), x) -> APP(*, x) APP(app(*, app(app(+, y), z)), x) -> APP(app(*, x), z) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, x), app(app(*, y), z)) APP(app(*, app(app(*, x), y)), z) -> APP(app(*, y), z) APP(app(*, app(app(*, x), y)), z) -> APP(*, y) APP(app(+, app(app(+, x), y)), z) -> APP(app(+, x), app(app(+, y), z)) APP(app(+, app(app(+, x), y)), z) -> APP(app(+, y), z) APP(app(+, app(app(+, x), y)), z) -> APP(+, 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) APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x)
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