details

property	value
status	complete
benchmark	torpa3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n075.star.cs.uiowa.edu
space	Secret_05_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.70363020897 seconds
cpu usage	18.362966739
max memory	2.772570112E9

stage attributes

key	value
output-size	4653
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 73 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(x1)) -> c(d(x1)) c(c(x1)) -> d(d(d(x1))) c(x1) -> g(x1) d(d(x1)) -> c(f(x1)) d(d(d(x1))) -> g(c(x1)) f(x1) -> a(g(x1)) g(x1) -> d(a(b(x1))) g(g(x1)) -> b(c(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: B(b(x1)) -> C(d(x1)) B(b(x1)) -> D(x1) C(c(x1)) -> D(d(d(x1))) C(c(x1)) -> D(d(x1)) C(c(x1)) -> D(x1) C(x1) -> G(x1) D(d(x1)) -> C(f(x1)) D(d(x1)) -> F(x1) D(d(d(x1))) -> G(c(x1)) D(d(d(x1))) -> C(x1) F(x1) -> G(x1) G(x1) -> D(a(b(x1))) G(x1) -> B(x1) G(g(x1)) -> B(c(x1)) G(g(x1)) -> C(x1) The TRS R consists of the following rules: b(b(x1)) -> c(d(x1)) c(c(x1)) -> d(d(d(x1))) c(x1) -> g(x1) d(d(x1)) -> c(f(x1)) d(d(d(x1))) -> g(c(x1)) f(x1) -> a(g(x1)) g(x1) -> d(a(b(x1))) g(g(x1)) -> b(c(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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: C(c(x1)) -> D(d(d(x1))) D(d(x1)) -> C(f(x1)) C(c(x1)) -> D(d(x1)) D(d(x1)) -> F(x1) F(x1) -> G(x1) G(x1) -> B(x1) B(b(x1)) -> C(d(x1)) popout

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

job log

popout

actions

all output return to SRS Stand 10685