Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
SRS Stand 10685 pair #381718564
details
property
value
status
complete
benchmark
3-matchbox.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n007.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
12.402712822 seconds
cpu usage
45.675658907
max memory
4.691812352E9
stage attributes
key
value
output-size
7315
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 19 ms] (4) QDP (5) MRRProof [EQUIVALENT, 105 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 0 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 26 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(b(a(a(x1)))) -> a(a(b(c(x1)))) b(a(a(a(x1)))) -> a(a(a(b(x1)))) a(b(c(x1))) -> c(b(a(x1))) c(c(b(b(x1)))) -> b(b(c(c(x1)))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(c(x1)))) -> c(b(a(a(x1)))) a(a(a(b(x1)))) -> b(a(a(a(x1)))) c(b(a(x1))) -> a(b(c(x1))) b(b(c(c(x1)))) -> c(c(b(b(x1)))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(b(c(x1)))) -> C(b(a(a(x1)))) A(a(b(c(x1)))) -> B(a(a(x1))) A(a(b(c(x1)))) -> A(a(x1)) A(a(b(c(x1)))) -> A(x1) A(a(a(b(x1)))) -> B(a(a(a(x1)))) A(a(a(b(x1)))) -> A(a(a(x1))) A(a(a(b(x1)))) -> A(a(x1)) A(a(a(b(x1)))) -> A(x1) C(b(a(x1))) -> A(b(c(x1))) C(b(a(x1))) -> B(c(x1)) C(b(a(x1))) -> C(x1) B(b(c(c(x1)))) -> C(c(b(b(x1)))) B(b(c(c(x1)))) -> C(b(b(x1))) B(b(c(c(x1)))) -> B(b(x1)) B(b(c(c(x1)))) -> B(x1) The TRS R consists of the following rules: a(a(b(c(x1)))) -> c(b(a(a(x1)))) a(a(a(b(x1)))) -> b(a(a(a(x1)))) c(b(a(x1))) -> a(b(c(x1))) b(b(c(c(x1)))) -> c(c(b(b(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to SRS Stand 10685