details

property	value
status	complete
benchmark	size-12-alpha-3-num-552.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n006.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	7.78625202179 seconds
cpu usage	27.772640112
max memory	3.922743296E9

stage attributes

key	value
output-size	3125
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 109 ms] (6) QDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(x1)) -> x1 a(c(x1)) -> b(c(c(a(a(b(x1)))))) b(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(c(x1)) -> B(c(c(a(a(b(x1)))))) A(c(x1)) -> A(a(b(x1))) A(c(x1)) -> A(b(x1)) A(c(x1)) -> B(x1) The TRS R consists of the following rules: a(b(x1)) -> x1 a(c(x1)) -> b(c(c(a(a(b(x1)))))) b(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(c(x1)) -> A(b(x1)) A(c(x1)) -> A(a(b(x1))) The TRS R consists of the following rules: a(b(x1)) -> x1 a(c(x1)) -> b(c(c(a(a(b(x1)))))) b(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. A(c(x1)) -> A(b(x1)) A(c(x1)) -> A(a(b(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO,RATPOLO]: POL(A(x_1)) = [1/2]x_1 POL(a(x_1)) = [4]x_1 POL(b(x_1)) = [1/4]x_1 POL(c(x_1)) = [1/2] + [4]x_1 popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to SRS Stand 10685