Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381714714
details
property
value
status
complete
benchmark
Ex15_Luc98_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n026.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.182009935379 seconds
cpu usage
0.178564671
max memory
4046848.0
stage attributes
key
value
output-size
13884
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. 0 : [] --> o a!6220!6220add : [o * o] --> o a!6220!6220and : [o * o] --> o a!6220!6220first : [o * o] --> o a!6220!6220from : [o] --> o a!6220!6220if : [o * o * o] --> o add : [o * o] --> o and : [o * o] --> o cons : [o * o] --> o false : [] --> o first : [o * o] --> o from : [o] --> o if : [o * o * o] --> o mark : [o] --> o nil : [] --> o s : [o] --> o true : [] --> o a!6220!6220and(true, X) => mark(X) a!6220!6220and(false, X) => false a!6220!6220if(true, X, Y) => mark(X) a!6220!6220if(false, X, Y) => mark(Y) a!6220!6220add(0, X) => mark(X) a!6220!6220add(s(X), Y) => s(add(X, Y)) a!6220!6220first(0, X) => nil a!6220!6220first(s(X), cons(Y, Z)) => cons(Y, first(X, Z)) a!6220!6220from(X) => cons(X, from(s(X))) mark(and(X, Y)) => a!6220!6220and(mark(X), Y) mark(if(X, Y, Z)) => a!6220!6220if(mark(X), Y, Z) mark(add(X, Y)) => a!6220!6220add(mark(X), Y) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(from(X)) => a!6220!6220from(X) mark(true) => true mark(false) => false mark(0) => 0 mark(s(X)) => s(X) mark(nil) => nil mark(cons(X, Y)) => cons(X, Y) a!6220!6220and(X, Y) => and(X, Y) a!6220!6220if(X, Y, Z) => if(X, Y, Z) a!6220!6220add(X, Y) => add(X, Y) a!6220!6220first(X, Y) => first(X, Y) a!6220!6220from(X) => from(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]): a!6220!6220and(true, X) >? mark(X) a!6220!6220and(false, X) >? false a!6220!6220if(true, X, Y) >? mark(X) a!6220!6220if(false, X, Y) >? mark(Y) a!6220!6220add(0, X) >? mark(X) a!6220!6220add(s(X), Y) >? s(add(X, Y)) a!6220!6220first(0, X) >? nil a!6220!6220first(s(X), cons(Y, Z)) >? cons(Y, first(X, Z)) a!6220!6220from(X) >? cons(X, from(s(X))) mark(and(X, Y)) >? a!6220!6220and(mark(X), Y) mark(if(X, Y, Z)) >? a!6220!6220if(mark(X), Y, Z) mark(add(X, Y)) >? a!6220!6220add(mark(X), Y) mark(first(X, Y)) >? a!6220!6220first(mark(X), mark(Y)) mark(from(X)) >? a!6220!6220from(X) mark(true) >? true mark(false) >? false mark(0) >? 0 mark(s(X)) >? s(X) mark(nil) >? nil mark(cons(X, Y)) >? cons(X, Y) a!6220!6220and(X, Y) >? and(X, Y) a!6220!6220if(X, Y, Z) >? if(X, Y, Z) a!6220!6220add(X, Y) >? add(X, Y) a!6220!6220first(X, Y) >? first(X, Y) a!6220!6220from(X) >? from(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220add = \y0y1.y0 + 2y1 a!6220!6220and = \y0y1.2y0 + 2y1 a!6220!6220first = \y0y1.y0 + y1 a!6220!6220from = \y0.3y0 a!6220!6220if = \y0y1y2.y0 + 2y1 + 2y2 add = \y0y1.y0 + 2y1 and = \y0y1.2y0 + 2y1 cons = \y0y1.y0 + y1 false = 2 first = \y0y1.y0 + y1
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472