details

property	value
status	complete
benchmark	test9.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n020.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.8413040638 seconds
cpu usage	4.092101198
max memory	2.97889792E8

stage attributes

key	value
output-size	3359
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) QDP (5) QReductionProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, X) -> h(X, X) h(0, X) -> f(0, X, X) g(X, Y) -> X g(X, Y) -> Y The set Q consists of the following terms: f(0, 1, x0) h(0, x0) g(x0, x1) ---------------------------------------- (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) -> H(X, X) H(0, X) -> F(0, X, X) The TRS R consists of the following rules: f(0, 1, X) -> h(X, X) h(0, X) -> f(0, X, X) g(X, Y) -> X g(X, Y) -> Y The set Q consists of the following terms: f(0, 1, x0) h(0, x0) g(x0, x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(0, 1, X) -> H(X, X) H(0, X) -> F(0, X, X) R is empty. The set Q consists of the following terms: f(0, 1, x0) h(0, x0) g(x0, x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. popout

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actions

all output return to TRS Inner 89993