details

property	value
status	complete
benchmark	Ex1_GM99.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n034.star.cs.uiowa.edu
space	Strategy_removed_CSR_05

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.34161996841 seconds
cpu usage	6.131852879
max memory	4.8361472E8

stage attributes

key	value
output-size	2172
starexec-result	NO

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof [COMPLETE, 0 ms] (4) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, b, X) -> f(X, X, X) c -> a c -> b 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: F(a, b, X) -> F(X, X, X) The TRS R consists of the following rules: f(a, b, X) -> f(X, X, X) c -> a c -> b Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = F(c, c, X) evaluates to t =F(X, X, X) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [X / c] -------------------------------------------------------------------------------- Rewriting sequence F(c, c, c) -> F(c, b, c) with rule c -> b at position [1] and matcher [ ] F(c, b, c) -> F(a, b, c) with rule c -> a at position [0] and matcher [ ] F(a, b, c) -> F(c, c, c) with rule F(a, b, X) -> F(X, X, X) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (4) NO popout

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

job log

popout

actions

all output return to TRS Standard