details

property	value
status	complete
benchmark	2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n034.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	6.72530293465 seconds
cpu usage	23.225776145
max memory	3.634434048E9

stage attributes

key	value
output-size	4188
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 3 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 20 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 43 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(x1))) -> a(a(b(x1))) b(a(b(x1))) -> 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(a(a(x1))) -> b(a(a(x1))) b(a(b(x1))) -> a(b(a(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(a(a(x1))) -> B(a(a(x1))) B(a(b(x1))) -> A(b(a(x1))) B(a(b(x1))) -> B(a(x1)) B(a(b(x1))) -> A(x1) The TRS R consists of the following rules: a(a(a(x1))) -> b(a(a(x1))) b(a(b(x1))) -> a(b(a(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(a(b(x1))) -> B(a(x1)) B(a(b(x1))) -> A(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: b(a(b(x1))) -> a(b(a(x1))) a(a(a(x1))) -> b(a(a(x1))) popout

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