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TRS Stand 20472 pair #381712872
details
property
value
status
complete
benchmark
028.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n008.star.cs.uiowa.edu
space
AotoYamada_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.6687579155 seconds
cpu usage
3.726122287
max memory
2.33504768E8
stage attributes
key
value
output-size
5565
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 29 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(app(consif, true), x), ys) -> app(app(cons, x), ys) app(app(app(consif, false), x), ys) -> ys app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(consif, app(f, x)), x), app(app(filter, f), xs)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(app(consif, true), x), ys) -> app(app(cons, x), ys) app(app(app(consif, false), x), ys) -> ys app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(consif, app(f, x)), x), app(app(filter, f), xs)) The set Q consists of the following terms: app(app(app(consif, true), x0), x1) app(app(app(consif, false), x0), x1) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(consif, true), x), ys) -> APP(app(cons, x), ys) APP(app(app(consif, true), x), ys) -> APP(cons, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(consif, app(f, x)), x), app(app(filter, f), xs)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(consif, app(f, x)), x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(consif, app(f, x)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs) The TRS R consists of the following rules: app(app(app(consif, true), x), ys) -> app(app(cons, x), ys) app(app(app(consif, false), x), ys) -> ys app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(consif, app(f, x)), x), app(app(filter, f), xs)) The set Q consists of the following terms: app(app(app(consif, true), x0), x1) app(app(app(consif, false), x0), x1) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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