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SRS Stand 10685 pair #381717390
details
property
value
status
complete
benchmark
10.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
Zantema_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
11.8705320358 seconds
cpu usage
31.842424818
max memory
4.164194304E9
stage attributes
key
value
output-size
5264
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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, 31 ms] (6) QDP (7) QDPOrderProof [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: a(d(x1)) -> d(b(x1)) a(x1) -> b(b(b(x1))) d(x1) -> x1 a(x1) -> x1 b(d(b(x1))) -> a(d(x1)) b(c(x1)) -> c(d(d(x1))) a(c(x1)) -> b(b(c(d(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(d(x1)) -> D(b(x1)) A(d(x1)) -> B(x1) A(x1) -> B(b(b(x1))) A(x1) -> B(b(x1)) A(x1) -> B(x1) B(d(b(x1))) -> A(d(x1)) B(d(b(x1))) -> D(x1) B(c(x1)) -> D(d(x1)) B(c(x1)) -> D(x1) A(c(x1)) -> B(b(c(d(x1)))) A(c(x1)) -> B(c(d(x1))) A(c(x1)) -> D(x1) The TRS R consists of the following rules: a(d(x1)) -> d(b(x1)) a(x1) -> b(b(b(x1))) d(x1) -> x1 a(x1) -> x1 b(d(b(x1))) -> a(d(x1)) b(c(x1)) -> c(d(d(x1))) a(c(x1)) -> b(b(c(d(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 6 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(d(x1)) -> B(x1) B(d(b(x1))) -> A(d(x1)) A(x1) -> B(b(b(x1))) A(x1) -> B(b(x1)) A(x1) -> B(x1) A(c(x1)) -> B(b(c(d(x1)))) The TRS R consists of the following rules: a(d(x1)) -> d(b(x1))
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