details

property	value
status	complete
benchmark	random-332.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	13.1189830303 seconds
cpu usage	48.89112203
max memory	1.989537792E9

stage attributes

key	value
output-size	8385
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 9 ms] (4) QDP (5) MRRProof [EQUIVALENT, 53 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 442 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 160 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 77 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(b(x1)))) -> b(a(b(a(x1)))) b(a(a(a(x1)))) -> b(a(b(b(x1)))) a(b(a(b(x1)))) -> b(b(b(b(x1)))) a(b(a(b(x1)))) -> a(a(a(b(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: b(a(a(b(x1)))) -> a(b(a(b(x1)))) a(a(a(b(x1)))) -> b(b(a(b(x1)))) b(a(b(a(x1)))) -> b(b(b(b(x1)))) b(a(b(a(x1)))) -> b(a(a(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: B(a(a(b(x1)))) -> A(b(a(b(x1)))) B(a(a(b(x1)))) -> B(a(b(x1))) A(a(a(b(x1)))) -> B(b(a(b(x1)))) A(a(a(b(x1)))) -> B(a(b(x1))) B(a(b(a(x1)))) -> B(b(b(b(x1)))) B(a(b(a(x1)))) -> B(b(b(x1))) B(a(b(a(x1)))) -> B(b(x1)) B(a(b(a(x1)))) -> B(x1) B(a(b(a(x1)))) -> B(a(a(a(x1)))) B(a(b(a(x1)))) -> A(a(a(x1))) B(a(b(a(x1)))) -> A(a(x1)) The TRS R consists of the following rules: b(a(a(b(x1)))) -> a(b(a(b(x1)))) a(a(a(b(x1)))) -> b(b(a(b(x1)))) b(a(b(a(x1)))) -> b(b(b(b(x1)))) b(a(b(a(x1)))) -> b(a(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. popout

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