details

property	value
status	complete
benchmark	cime4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n062.star.cs.uiowa.edu
space	Secret_05_TRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.46791195869 seconds
cpu usage	6.484928995
max memory	4.84995072E8

stage attributes

key	value
output-size	3571
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) NonTerminationLoopProof [COMPLETE, 0 ms] (6) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(X) -> u(h(X), h(X), X) u(d, c(Y), X) -> k(Y) h(d) -> c(a) h(d) -> c(b) f(k(a), k(b), X) -> f(X, X, 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: G(X) -> U(h(X), h(X), X) G(X) -> H(X) F(k(a), k(b), X) -> F(X, X, X) The TRS R consists of the following rules: g(X) -> u(h(X), h(X), X) u(d, c(Y), X) -> k(Y) h(d) -> c(a) h(d) -> c(b) f(k(a), k(b), X) -> f(X, X, 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 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(k(a), k(b), X) -> F(X, X, X) The TRS R consists of the following rules: g(X) -> u(h(X), h(X), X) u(d, c(Y), X) -> k(Y) h(d) -> c(a) h(d) -> c(b) f(k(a), k(b), X) -> f(X, X, X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) 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(u(d, h(d), X'), u(d, h(d), X''), 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'' / X', X / u(d, h(d), X')] -------------------------------------------------------------------------------- Rewriting sequence popout

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

job log

popout

actions

all output return to TRS Standard