details

property	value
status	complete
benchmark	5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	11.9361600876 seconds
cpu usage	44.079773438
max memory	4.400422912E9

stage attributes

key	value
output-size	10786
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 15 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 1 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 217 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 56 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 249 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 120 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 104 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 887 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(x1))) -> a(a(b(b(x1)))) b(b(b(b(x1)))) -> a(b(a(a(x1)))) 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: A(a(a(x1))) -> A(a(b(b(x1)))) A(a(a(x1))) -> A(b(b(x1))) A(a(a(x1))) -> B(b(x1)) A(a(a(x1))) -> B(x1) B(b(b(b(x1)))) -> A(b(a(a(x1)))) B(b(b(b(x1)))) -> B(a(a(x1))) B(b(b(b(x1)))) -> A(a(x1)) B(b(b(b(x1)))) -> A(x1) The TRS R consists of the following rules: a(a(a(x1))) -> a(a(b(b(x1)))) b(b(b(b(x1)))) -> a(b(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(a(x1))) -> A(b(b(x1))) A(a(a(x1))) -> A(a(b(b(x1)))) A(a(a(x1))) -> B(b(x1)) B(b(b(b(x1)))) -> A(b(a(a(x1)))) A(a(a(x1))) -> B(x1) B(b(b(b(x1)))) -> A(a(x1)) B(b(b(b(x1)))) -> A(x1) The TRS R consists of the following rules: a(a(a(x1))) -> a(a(b(b(x1)))) b(b(b(b(x1)))) -> a(b(a(a(x1)))) Q is empty. popout

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