details

property	value
status	complete
benchmark	ex2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n009.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.03115177155 seconds
cpu usage	4.739171015
max memory	3.44346624E8

stage attributes

key	value
output-size	4140
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) NonTerminationLoopProof [COMPLETE, 0 ms] (12) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: cons(x, cons(y, z)) -> big inf(x) -> cons(x, inf(s(x))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: INF(x) -> CONS(x, inf(s(x))) INF(x) -> INF(s(x)) The TRS R consists of the following rules: cons(x, cons(y, z)) -> big inf(x) -> cons(x, inf(s(x))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: INF(x) -> INF(s(x)) The TRS R consists of the following rules: cons(x, cons(y, z)) -> big inf(x) -> cons(x, inf(s(x))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: INF(x) -> INF(s(x)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) TransformationProof (EQUIVALENT) popout

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

job log

popout

actions

all output return to TRS Stand 20472