details

property	value
status	complete
benchmark	2.46.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n190.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.9414 seconds
cpu usage	8.34051
user time	7.96104
system time	0.379477
max virtual memory	1.9874796E7
max residence set size	969588.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 39 ms] (2) QTRS (3) Overlay + Local Confluence [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 2 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 29 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(x)) -> b(b(x)) b(b(a(x))) -> a(b(b(x))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a(a(x)) -> b(b(x)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(a(x))) -> a(b(b(x))) Q is empty. ---------------------------------------- (3) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(a(x))) -> a(b(b(x))) The set Q consists of the following terms: b(b(a(x0))) ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(a(x))) -> B(b(x)) B(b(a(x))) -> B(x) The TRS R consists of the following rules: b(b(a(x))) -> a(b(b(x))) The set Q consists of the following terms: b(b(a(x0))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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job log

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actions

all output return to TRS Standard