details

property	value
status	complete
benchmark	#4.13.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n043.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	2.58505296707 seconds
cpu usage	7.108442915
max memory	4.635648E8

stage attributes

key	value
output-size	2714
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof [COMPLETE, 707 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) f(x, y, z) -> 2 0 -> 2 1 -> 2 g(x, x, y) -> y g(x, y, y) -> 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: F(0, 1, x) -> F(x, x, x) The TRS R consists of the following rules: f(0, 1, x) -> f(x, x, x) f(x, y, z) -> 2 0 -> 2 1 -> 2 g(x, x, y) -> y g(x, y, y) -> x 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(1, 0, 0), g(1, y, y), 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 / 0, x / g(1, 0, 0)] -------------------------------------------------------------------------------- Rewriting sequence F(g(1, 0, 0), g(1, 0, 0), g(1, 0, 0)) -> F(g(1, 0, 0), 1, g(1, 0, 0)) with rule g(x', y, y) -> x' at position [1] and matcher [x' / 1, y / 0] F(g(1, 0, 0), 1, g(1, 0, 0)) -> F(g(1, 2, 0), 1, g(1, 0, 0)) with rule 0 -> 2 at position [0,1] and matcher [ ] F(g(1, 2, 0), 1, g(1, 0, 0)) -> F(g(2, 2, 0), 1, g(1, 0, 0)) with rule 1 -> 2 at position [0,0] and matcher [ ] F(g(2, 2, 0), 1, g(1, 0, 0)) -> F(0, 1, g(1, 0, 0)) with rule g(x', x', y) -> y at position [0] and matcher [x' / 2, y / 0] F(0, 1, g(1, 0, 0)) -> F(g(1, 0, 0), g(1, 0, 0), g(1, 0, 0)) 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. ---------------------------------------- popout

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actions

all output return to TRS Standard