Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381714964
details
property
value
status
complete
benchmark
Ex3_3_25_Bor03_L.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n108.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0262849330902 seconds
cpu usage
0.020863211
max memory
1396736.0
stage attributes
key
value
output-size
2267
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. app : [o * o] --> o cons : [o] --> o from : [o] --> o nil : [] --> o prefix : [o] --> o zWadr : [o * o] --> o app(nil, X) => X app(cons(X), Y) => cons(X) from(X) => cons(X) zWadr(nil, X) => nil zWadr(X, nil) => nil zWadr(cons(X), cons(Y)) => cons(app(Y, cons(X))) prefix(X) => cons(nil) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(nil, X) >? X app(cons(X), Y) >? cons(X) from(X) >? cons(X) zWadr(nil, X) >? nil zWadr(X, nil) >? nil zWadr(cons(X), cons(Y)) >? cons(app(Y, cons(X))) prefix(X) >? cons(nil) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: app = \y0y1.2 + y0 + y1 cons = \y0.2 + 2y0 from = \y0.3 + 3y0 nil = 0 prefix = \y0.3 + 2y0 zWadr = \y0y1.3 + 3y0 + 3y1 Using this interpretation, the requirements translate to: [[app(nil, _x0)]] = 2 + x0 > x0 = [[_x0]] [[app(cons(_x0), _x1)]] = 4 + x1 + 2x0 > 2 + 2x0 = [[cons(_x0)]] [[from(_x0)]] = 3 + 3x0 > 2 + 2x0 = [[cons(_x0)]] [[zWadr(nil, _x0)]] = 3 + 3x0 > 0 = [[nil]] [[zWadr(_x0, nil)]] = 3 + 3x0 > 0 = [[nil]] [[zWadr(cons(_x0), cons(_x1))]] = 15 + 6x0 + 6x1 > 10 + 2x1 + 4x0 = [[cons(app(_x1, cons(_x0)))]] [[prefix(_x0)]] = 3 + 2x0 > 2 = [[cons(nil)]] We can thus remove the following rules: app(nil, X) => X app(cons(X), Y) => cons(X) from(X) => cons(X) zWadr(nil, X) => nil zWadr(X, nil) => nil zWadr(cons(X), cons(Y)) => cons(app(Y, cons(X))) prefix(X) => cons(nil) 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