details

property	value
status	complete
benchmark	random-361.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n075.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.92326283455 seconds
cpu usage	12.114332861
max memory	1.547378688E9

stage attributes

key	value
output-size	4788
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 13 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 5 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 59 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(b(a(x1)))) -> b(b(a(a(x1)))) a(a(a(b(x1)))) -> a(b(a(a(x1)))) a(a(b(a(x1)))) -> b(b(a(a(x1)))) a(b(b(b(x1)))) -> b(a(b(a(x1)))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(x1)))) -> a(a(b(b(x1)))) b(a(a(a(x1)))) -> a(a(b(a(x1)))) a(b(a(a(x1)))) -> a(a(b(b(x1)))) b(b(b(a(x1)))) -> a(b(a(b(x1)))) Q is empty. ---------------------------------------- (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: A(b(a(b(x1)))) -> A(a(b(b(x1)))) A(b(a(b(x1)))) -> A(b(b(x1))) A(b(a(b(x1)))) -> B(b(x1)) B(a(a(a(x1)))) -> A(a(b(a(x1)))) B(a(a(a(x1)))) -> A(b(a(x1))) B(a(a(a(x1)))) -> B(a(x1)) A(b(a(a(x1)))) -> A(a(b(b(x1)))) A(b(a(a(x1)))) -> A(b(b(x1))) A(b(a(a(x1)))) -> B(b(x1)) A(b(a(a(x1)))) -> B(x1) B(b(b(a(x1)))) -> A(b(a(b(x1)))) B(b(b(a(x1)))) -> B(a(b(x1))) B(b(b(a(x1)))) -> A(b(x1)) B(b(b(a(x1)))) -> B(x1) The TRS R consists of the following rules: a(b(a(b(x1)))) -> a(a(b(b(x1)))) b(a(a(a(x1)))) -> a(a(b(a(x1)))) a(b(a(a(x1)))) -> a(a(b(b(x1)))) b(b(b(a(x1)))) -> a(b(a(b(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) popout

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