details

property	value
status	complete
benchmark	#4.25.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n183.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	3.42442202568 seconds
cpu usage	5.464607793
max memory	4.75222016E8

stage attributes

key	value
output-size	3930
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, x) -> f(g(x), x) g(x) -> s(x) The set Q consists of the following terms: f(x0, x0) g(x0) ---------------------------------------- (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: F(x, x) -> F(g(x), x) F(x, x) -> G(x) The TRS R consists of the following rules: f(x, x) -> f(g(x), x) g(x) -> s(x) The set Q consists of the following terms: f(x0, x0) g(x0) 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: F(x, x) -> F(g(x), x) The TRS R consists of the following rules: f(x, x) -> f(g(x), x) g(x) -> s(x) The set Q consists of the following terms: f(x0, x0) g(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) popout

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

job log

popout

actions

all output return to TRS Innermost