Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381712140
details
property
value
status
complete
benchmark
07.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n087.star.cs.uiowa.edu
space
Der95
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0227048397064 seconds
cpu usage
0.019495837
max memory
1396736.0
stage attributes
key
value
output-size
1463
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. b : [o] --> o r : [o] --> o w : [o] --> o w(r(X)) => r(w(X)) b(r(X)) => r(b(X)) b(w(X)) => w(b(X)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): w(r(X)) >? r(w(X)) b(r(X)) >? r(b(X)) b(w(X)) >? w(b(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: b = \y0.2 + 3y0 r = \y0.2 + y0 w = \y0.3 + 2y0 Using this interpretation, the requirements translate to: [[w(r(_x0))]] = 7 + 2x0 > 5 + 2x0 = [[r(w(_x0))]] [[b(r(_x0))]] = 8 + 3x0 > 4 + 3x0 = [[r(b(_x0))]] [[b(w(_x0))]] = 11 + 6x0 > 7 + 6x0 = [[w(b(_x0))]] We can thus remove the following rules: w(r(X)) => r(w(X)) b(r(X)) => r(b(X)) b(w(X)) => w(b(X)) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472