details

property	value
status	complete
benchmark	03.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n037.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	10.7032139301 seconds
cpu usage	39.491201791
max memory	4.167675904E9

stage attributes

key	value
output-size	3797
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 8 ms] (4) QDP (5) MRRProof [EQUIVALENT, 20 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 421 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(b(x1)))) -> b(b(x1)) b(b(x1)) -> a(b(a(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: b(a(a(a(x1)))) -> b(b(x1)) b(b(x1)) -> a(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: B(a(a(a(x1)))) -> B(b(x1)) B(a(a(a(x1)))) -> B(x1) B(b(x1)) -> B(a(x1)) The TRS R consists of the following rules: b(a(a(a(x1)))) -> b(b(x1)) b(b(x1)) -> a(a(b(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. Strictly oriented dependency pairs: B(a(a(a(x1)))) -> B(x1) B(b(x1)) -> B(a(x1)) Used ordering: Polynomial interpretation [POLO]: POL(B(x_1)) = x_1 POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 3 + x_1 ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to SRS Stand 10685