details

property	value
status	complete
benchmark	#4.12a.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.83851909637 seconds
cpu usage	4.078319306
max memory	3.03075328E8

stage attributes

key	value
output-size	2533
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof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: h(x, y) -> f(x, y, x) f(0, 1, x) -> h(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: H(x, y) -> F(x, y, x) F(0, 1, x) -> H(x, x) The TRS R consists of the following rules: h(x, y) -> f(x, y, x) f(0, 1, x) -> h(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)) -> H(g(0, 1), g(0, 1)) with rule F(0, 1, x') -> H(x', x') at position [] and matcher [x' / g(0, 1)] H(g(0, 1), g(0, 1)) -> F(g(0, 1), g(0, 1), g(0, 1)) with rule H(x, y) -> F(x, y, 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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