details

property	value
status	complete
benchmark	jw1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n177.star.cs.uiowa.edu
space	Waldmann_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.66245698929 seconds
cpu usage	15.251793871
max memory	1.95561472E9

stage attributes

key	value
output-size	3914
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 1 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 30 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 160 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(b(x1))) -> a(a(a(x1))) a(a(a(x1))) -> b(a(b(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: B(b(b(x1))) -> A(a(a(x1))) B(b(b(x1))) -> A(a(x1)) B(b(b(x1))) -> A(x1) A(a(a(x1))) -> B(a(b(x1))) A(a(a(x1))) -> A(b(x1)) A(a(a(x1))) -> B(x1) The TRS R consists of the following rules: b(b(b(x1))) -> a(a(a(x1))) a(a(a(x1))) -> b(a(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(b(b(x1))) -> A(a(x1)) B(b(b(x1))) -> A(x1) A(a(a(x1))) -> A(b(x1)) A(a(a(x1))) -> B(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = x_1 POL(B(x_1)) = x_1 POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: a(a(a(x1))) -> b(a(b(x1))) b(b(b(x1))) -> a(a(a(x1))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(b(x1))) -> A(a(a(x1))) A(a(a(x1))) -> B(a(b(x1))) The TRS R consists of the following rules: b(b(b(x1))) -> a(a(a(x1))) popout

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