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TRS Stand 20472 pair #381712550
details
property
value
status
complete
benchmark
29.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n084.star.cs.uiowa.edu
space
Applicative_first_order_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.15509009361 seconds
cpu usage
5.229098013
max memory
2.76414464E8
stage attributes
key
value
output-size
12227
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) QDPApplicativeOrderProof [EQUIVALENT, 0 ms] (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) ATransformationProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(ack, 0), y) -> app(succ, y) app(app(ack, app(succ, x)), y) -> app(app(ack, x), app(succ, 0)) app(app(ack, app(succ, x)), app(succ, y)) -> app(app(ack, x), app(app(ack, app(succ, 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) 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(ack, 0), y) -> APP(succ, y) APP(app(ack, app(succ, x)), y) -> APP(app(ack, x), app(succ, 0)) APP(app(ack, app(succ, x)), y) -> APP(ack, x) APP(app(ack, app(succ, x)), y) -> APP(succ, 0) APP(app(ack, app(succ, x)), app(succ, y)) -> APP(app(ack, x), app(app(ack, app(succ, x)), y)) APP(app(ack, app(succ, x)), app(succ, y)) -> APP(ack, x) APP(app(ack, app(succ, x)), app(succ, y)) -> APP(app(ack, app(succ, x)), 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) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) The TRS R consists of the following rules: app(app(ack, 0), y) -> app(succ, y) app(app(ack, app(succ, x)), y) -> app(app(ack, x), app(succ, 0)) app(app(ack, app(succ, x)), app(succ, y)) -> app(app(ack, x), app(app(ack, app(succ, 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) Q is empty.
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