Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
SRS Stand 10685 pair #381717182
details
property
value
status
complete
benchmark
hom01.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n086.star.cs.uiowa.edu
space
Trafo_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
9.28949999809 seconds
cpu usage
33.538799744
max memory
4.175986688E9
stage attributes
key
value
output-size
6520
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 19 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 557 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 4 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(d(b(d(a(x1))))))) -> a(a(c(a(a(b(d(x1))))))) a(a(c(x1))) -> c(c(a(a(x1)))) c(c(c(x1))) -> b(d(c(b(d(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(d(b(d(b(a(a(x1))))))) -> d(b(a(a(c(a(a(x1))))))) c(a(a(x1))) -> a(a(c(c(x1)))) c(c(c(x1))) -> d(b(c(d(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(d(b(d(b(a(a(x1))))))) -> A(a(c(a(a(x1))))) A(d(b(d(b(a(a(x1))))))) -> A(c(a(a(x1)))) A(d(b(d(b(a(a(x1))))))) -> C(a(a(x1))) C(a(a(x1))) -> A(a(c(c(x1)))) C(a(a(x1))) -> A(c(c(x1))) C(a(a(x1))) -> C(c(x1)) C(a(a(x1))) -> C(x1) C(c(c(x1))) -> C(d(b(x1))) The TRS R consists of the following rules: a(d(b(d(b(a(a(x1))))))) -> d(b(a(a(c(a(a(x1))))))) c(a(a(x1))) -> a(a(c(c(x1)))) c(c(c(x1))) -> d(b(c(d(b(x1))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(d(b(d(b(a(a(x1))))))) -> A(c(a(a(x1))))
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to SRS Stand 10685