details

property	value
status	complete
benchmark	size-12-alpha-3-num-288.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n073.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	6.50713300705 seconds
cpu usage	22.498343798
max memory	1.471049728E9

stage attributes

key	value
output-size	3744
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 218 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> x1 a(b(x1)) -> c(b(c(b(a(x1))))) b(x1) -> a(x1) c(c(x1)) -> x1 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: A(b(x1)) -> C(b(c(b(a(x1))))) A(b(x1)) -> B(c(b(a(x1)))) A(b(x1)) -> C(b(a(x1))) A(b(x1)) -> B(a(x1)) A(b(x1)) -> A(x1) B(x1) -> A(x1) The TRS R consists of the following rules: a(x1) -> x1 a(b(x1)) -> c(b(c(b(a(x1))))) b(x1) -> a(x1) c(c(x1)) -> x1 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: A(b(x1)) -> B(c(b(a(x1)))) B(x1) -> A(x1) A(b(x1)) -> B(a(x1)) A(b(x1)) -> A(x1) The TRS R consists of the following rules: a(x1) -> x1 a(b(x1)) -> c(b(c(b(a(x1))))) b(x1) -> a(x1) c(c(x1)) -> x1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(x1) -> A(x1) A(b(x1)) -> A(x1) popout

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actions

all output return to SRS Standard