details

property	value
status	complete
benchmark	Ex5_DLMMU04_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	1.27431917191 seconds
cpu usage	1.271501888
max memory	1.8784256E7

stage attributes

key	value
output-size	54434
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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!6220incr : [o] --> o a!6220!6220oddNs : [] --> o a!6220!6220pairNs : [] --> o a!6220!6220repItems : [o] --> o a!6220!6220tail : [o] --> o a!6220!6220take : [o * o] --> o a!6220!6220zip : [o * o] --> o cons : [o * o] --> o incr : [o] --> o mark : [o] --> o nil : [] --> o oddNs : [] --> o pair : [o * o] --> o pairNs : [] --> o repItems : [o] --> o s : [o] --> o tail : [o] --> o take : [o * o] --> o zip : [o * o] --> o a!6220!6220pairNs => cons(0, incr(oddNs)) a!6220!6220oddNs => a!6220!6220incr(a!6220!6220pairNs) a!6220!6220incr(cons(X, Y)) => cons(s(mark(X)), incr(Y)) a!6220!6220take(0, X) => nil a!6220!6220take(s(X), cons(Y, Z)) => cons(mark(Y), take(X, Z)) a!6220!6220zip(nil, X) => nil a!6220!6220zip(X, nil) => nil a!6220!6220zip(cons(X, Y), cons(Z, U)) => cons(pair(mark(X), mark(Z)), zip(Y, U)) a!6220!6220tail(cons(X, Y)) => mark(Y) a!6220!6220repItems(nil) => nil a!6220!6220repItems(cons(X, Y)) => cons(mark(X), cons(X, repItems(Y))) mark(pairNs) => a!6220!6220pairNs mark(incr(X)) => a!6220!6220incr(mark(X)) mark(oddNs) => a!6220!6220oddNs mark(take(X, Y)) => a!6220!6220take(mark(X), mark(Y)) mark(zip(X, Y)) => a!6220!6220zip(mark(X), mark(Y)) mark(tail(X)) => a!6220!6220tail(mark(X)) mark(repItems(X)) => a!6220!6220repItems(mark(X)) mark(cons(X, Y)) => cons(mark(X), Y) mark(0) => 0 mark(s(X)) => s(mark(X)) mark(nil) => nil mark(pair(X, Y)) => pair(mark(X), mark(Y)) a!6220!6220pairNs => pairNs a!6220!6220incr(X) => incr(X) a!6220!6220oddNs => oddNs a!6220!6220take(X, Y) => take(X, Y) a!6220!6220zip(X, Y) => zip(X, Y) a!6220!6220tail(X) => tail(X) a!6220!6220repItems(X) => repItems(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!6220pairNs >? cons(0, incr(oddNs)) a!6220!6220oddNs >? a!6220!6220incr(a!6220!6220pairNs) a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) a!6220!6220take(0, X) >? nil a!6220!6220take(s(X), cons(Y, Z)) >? cons(mark(Y), take(X, Z)) a!6220!6220zip(nil, X) >? nil a!6220!6220zip(X, nil) >? nil a!6220!6220zip(cons(X, Y), cons(Z, U)) >? cons(pair(mark(X), mark(Z)), zip(Y, U)) a!6220!6220tail(cons(X, Y)) >? mark(Y) a!6220!6220repItems(nil) >? nil a!6220!6220repItems(cons(X, Y)) >? cons(mark(X), cons(X, repItems(Y))) mark(pairNs) >? a!6220!6220pairNs mark(incr(X)) >? a!6220!6220incr(mark(X)) mark(oddNs) >? a!6220!6220oddNs mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) mark(zip(X, Y)) >? a!6220!6220zip(mark(X), mark(Y)) mark(tail(X)) >? a!6220!6220tail(mark(X)) mark(repItems(X)) >? a!6220!6220repItems(mark(X)) mark(cons(X, Y)) >? cons(mark(X), Y) mark(0) >? 0 mark(s(X)) >? s(mark(X)) mark(nil) >? nil mark(pair(X, Y)) >? pair(mark(X), mark(Y)) a!6220!6220pairNs >? pairNs a!6220!6220incr(X) >? incr(X) a!6220!6220oddNs >? oddNs a!6220!6220take(X, Y) >? take(X, Y) a!6220!6220zip(X, Y) >? zip(X, Y) a!6220!6220tail(X) >? tail(X) a!6220!6220repItems(X) >? repItems(X) We orient these requirements with a polynomial interpretation in the natural numbers. popout

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