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	n118.star.cs.uiowa.edu
space	Secret_07_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.20756 seconds
cpu usage	13.2698
user time	12.528
system time	0.741706
max virtual memory	2.0871976E7
max residence set size	2345284.0

stage attributes

key	value
starexec-result	YES

output

12.80/4.08 YES 13.12/4.14 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 13.12/4.14 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 13.12/4.14 13.12/4.14 13.12/4.14 Termination w.r.t. Q of the given QTRS could be proven: 13.12/4.14 13.12/4.14 (0) QTRS 13.12/4.14 (1) QTRSRRRProof [EQUIVALENT, 13 ms] 13.12/4.14 (2) QTRS 13.12/4.14 (3) DependencyPairsProof [EQUIVALENT, 32 ms] 13.12/4.14 (4) QDP 13.12/4.14 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 13.12/4.14 (6) QDP 13.12/4.14 (7) UsableRulesProof [EQUIVALENT, 0 ms] 13.12/4.14 (8) QDP 13.12/4.14 (9) QDPSizeChangeProof [EQUIVALENT, 1 ms] 13.12/4.14 (10) YES 13.12/4.14 13.12/4.14 13.12/4.14 ---------------------------------------- 13.12/4.14 13.12/4.14 (0) 13.12/4.14 Obligation: 13.12/4.14 Q restricted rewrite system: 13.12/4.14 The TRS R consists of the following rules: 13.12/4.14 13.12/4.14 1(0(x1)) -> 0(0(0(1(x1)))) 13.12/4.14 0(1(x1)) -> 1(x1) 13.12/4.14 1(1(x1)) -> 0(0(0(0(x1)))) 13.12/4.14 0(0(x1)) -> 0(x1) 13.12/4.14 13.12/4.14 Q is empty. 13.12/4.14 13.12/4.14 ---------------------------------------- 13.12/4.14 13.12/4.14 (1) QTRSRRRProof (EQUIVALENT) 13.12/4.14 Used ordering: 13.12/4.14 Polynomial interpretation [POLO]: 13.12/4.14 13.12/4.14 POL(0(x_1)) = x_1 13.12/4.14 POL(1(x_1)) = 1 + x_1 13.12/4.14 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 13.12/4.14 13.12/4.14 1(1(x1)) -> 0(0(0(0(x1)))) 13.12/4.14 13.12/4.14 13.12/4.14 13.12/4.14 13.12/4.14 ---------------------------------------- 13.12/4.14 13.12/4.14 (2) 13.12/4.14 Obligation: 13.12/4.14 Q restricted rewrite system: 13.12/4.14 The TRS R consists of the following rules: 13.12/4.14 13.12/4.14 1(0(x1)) -> 0(0(0(1(x1)))) 13.12/4.14 0(1(x1)) -> 1(x1) 13.12/4.14 0(0(x1)) -> 0(x1) 13.12/4.14 13.12/4.14 Q is empty. 13.12/4.14 13.12/4.14 ---------------------------------------- 13.12/4.14 13.12/4.14 (3) DependencyPairsProof (EQUIVALENT) 13.12/4.14 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 13.12/4.14 ---------------------------------------- 13.12/4.14 13.12/4.14 (4) 13.12/4.14 Obligation: 13.12/4.14 Q DP problem: 13.12/4.14 The TRS P consists of the following rules: 13.12/4.14 13.12/4.14 1^1(0(x1)) -> 0^1(0(0(1(x1)))) 13.12/4.14 1^1(0(x1)) -> 0^1(0(1(x1))) 13.12/4.14 1^1(0(x1)) -> 0^1(1(x1)) 13.12/4.14 1^1(0(x1)) -> 1^1(x1) 13.12/4.14 13.12/4.14 The TRS R consists of the following rules: 13.12/4.14 13.12/4.14 1(0(x1)) -> 0(0(0(1(x1)))) 13.12/4.14 0(1(x1)) -> 1(x1) 13.12/4.14 0(0(x1)) -> 0(x1) 13.12/4.14 13.12/4.14 Q is empty. 13.12/4.14 We have to consider all minimal (P,Q,R)-chains. 13.12/4.14 ---------------------------------------- 13.12/4.14 13.12/4.14 (5) DependencyGraphProof (EQUIVALENT) 13.12/4.14 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. 13.12/4.14 ---------------------------------------- 13.12/4.14 13.12/4.14 (6) 13.12/4.14 Obligation: 13.12/4.14 Q DP problem: 13.12/4.14 The TRS P consists of the following rules: 13.12/4.14 13.12/4.14 1^1(0(x1)) -> 1^1(x1) 13.12/4.14 13.12/4.14 The TRS R consists of the following rules: popout

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

job log

popout

actions

all output return to SRS_Standard 2019-03-29 03.29