details

property	value
status	complete
benchmark	dj.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n010.star.cs.uiowa.edu
space	Secret_07_SRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	6.6497220993 seconds
cpu usage	12.279802699
max memory	1.220722688E9

stage attributes

key	value
output-size	3561
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 65 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 13 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 1(1(x1)) -> 0(0(0(0(x1)))) 0(0(x1)) -> 0(x1) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 1(1(x1)) -> 0(0(0(0(x1)))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 0(0(x1)) -> 0(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: 1^1(0(x1)) -> 0^1(0(0(1(x1)))) 1^1(0(x1)) -> 0^1(0(1(x1))) 1^1(0(x1)) -> 0^1(1(x1)) 1^1(0(x1)) -> 1^1(x1) The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 0(0(x1)) -> 0(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) Obligation: popout

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

job log

popout

actions

all output return to SRS Standard