details

property	value
status	complete
benchmark	size-12-alpha-3-num-545.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n030.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	6.83858895302 seconds
cpu usage	22.841290804
max memory	2.717642752E9

stage attributes

key	value
output-size	4139
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 5 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 48 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 110 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(x1)) -> x1 a(c(x1)) -> a(b(b(c(c(a(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)) -> A(b(b(c(c(a(x1)))))) A(c(x1)) -> B(b(c(c(a(x1))))) A(c(x1)) -> B(c(c(a(x1)))) A(c(x1)) -> A(x1) The TRS R consists of the following rules: a(b(x1)) -> x1 a(c(x1)) -> a(b(b(c(c(a(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(x1) A(c(x1)) -> A(b(b(c(c(a(x1)))))) The TRS R consists of the following rules: a(b(x1)) -> x1 a(c(x1)) -> a(b(b(c(c(a(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(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A_1(x_1) ) = x_1 POL( b_1(x_1) ) = max{0, x_1 - 1} POL( c_1(x_1) ) = x_1 + 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