details

property	value
status	complete
benchmark	toyama.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n168.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	2.43276715279 seconds
cpu usage	4.532669012
max memory	2.66166272E8

stage attributes

key	value
output-size	2352
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof [COMPLETE,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(0, 1, X) -> f(X, X, X) g(X, Y) -> X g(X, Y) -> Y 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(0, 1, X) -> F(X, X, X) The TRS R consists of the following rules: f(0, 1, X) -> f(X, X, X) g(X, Y) -> X g(X, Y) -> Y 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(g(0, Y), g(X', 1), 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: [Y / 1, X' / 0, X / g(0, 1)] -------------------------------------------------------------------------------- Rewriting sequence F(g(0, 1), g(0, 1), g(0, 1)) -> F(g(0, 1), 1, g(0, 1)) with rule g(X', Y') -> Y' at position [1] and matcher [X' / 0, Y' / 1] F(g(0, 1), 1, g(0, 1)) -> F(0, 1, g(0, 1)) with rule g(X', Y) -> X' at position [0] and matcher [X' / 0, Y / 1] F(0, 1, g(0, 1)) -> F(g(0, 1), g(0, 1), g(0, 1)) with rule F(0, 1, 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

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all output return to TRS Standard