TRS Stand 20472 pair #381715785

details

property	value
status	complete
benchmark	ExIntrod_GM04_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.85303401947 seconds
cpu usage	0.827059872
max memory	1.9185664E7

stage attributes

key	value
output-size	33673
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  0 : [] --> o 
  active : [o] --> o 
  adx : [o] --> o 
  cons : [o * o] --> o 
  hd : [o] --> o 
  incr : [o] --> o 
  mark : [o] --> o 
  nats : [] --> o 
  ok : [o] --> o 
  proper : [o] --> o 
  s : [o] --> o 
  tl : [o] --> o 
  top : [o] --> o 
  zeros : [] --> o 

  active(nats) => mark(adx(zeros)) 
  active(zeros) => mark(cons(0, zeros)) 
  active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) 
  active(adx(cons(X, Y))) => mark(incr(cons(X, adx(Y)))) 
  active(hd(cons(X, Y))) => mark(X) 
  active(tl(cons(X, Y))) => mark(Y) 
  active(adx(X)) => adx(active(X)) 
  active(incr(X)) => incr(active(X)) 
  active(hd(X)) => hd(active(X)) 
  active(tl(X)) => tl(active(X)) 
  adx(mark(X)) => mark(adx(X)) 
  incr(mark(X)) => mark(incr(X)) 
  hd(mark(X)) => mark(hd(X)) 
  tl(mark(X)) => mark(tl(X)) 
  proper(nats) => ok(nats) 
  proper(adx(X)) => adx(proper(X)) 
  proper(zeros) => ok(zeros) 
  proper(cons(X, Y)) => cons(proper(X), proper(Y)) 
  proper(0) => ok(0) 
  proper(incr(X)) => incr(proper(X)) 
  proper(s(X)) => s(proper(X)) 
  proper(hd(X)) => hd(proper(X)) 
  proper(tl(X)) => tl(proper(X)) 
  adx(ok(X)) => ok(adx(X)) 
  cons(ok(X), ok(Y)) => ok(cons(X, Y)) 
  incr(ok(X)) => ok(incr(X)) 
  s(ok(X)) => ok(s(X)) 
  hd(ok(X)) => ok(hd(X)) 
  tl(ok(X)) => ok(tl(X)) 
  top(mark(X)) => top(proper(X)) 
  top(ok(X)) => top(active(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]):

  active(nats) >? mark(adx(zeros)) 
  active(zeros) >? mark(cons(0, zeros)) 
  active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) 
  active(adx(cons(X, Y))) >? mark(incr(cons(X, adx(Y)))) 
  active(hd(cons(X, Y))) >? mark(X) 
  active(tl(cons(X, Y))) >? mark(Y) 
  active(adx(X)) >? adx(active(X)) 
  active(incr(X)) >? incr(active(X)) 
  active(hd(X)) >? hd(active(X)) 
  active(tl(X)) >? tl(active(X)) 
  adx(mark(X)) >? mark(adx(X)) 
  incr(mark(X)) >? mark(incr(X)) 
  hd(mark(X)) >? mark(hd(X)) 
  tl(mark(X)) >? mark(tl(X)) 
  proper(nats) >? ok(nats) 
  proper(adx(X)) >? adx(proper(X)) 
  proper(zeros) >? ok(zeros) 
  proper(cons(X, Y)) >? cons(proper(X), proper(Y)) 
  proper(0) >? ok(0) 
  proper(incr(X)) >? incr(proper(X)) 
  proper(s(X)) >? s(proper(X)) 
  proper(hd(X)) >? hd(proper(X)) 
  proper(tl(X)) >? tl(proper(X)) 
  adx(ok(X)) >? ok(adx(X)) 
  cons(ok(X), ok(Y)) >? ok(cons(X, Y)) 
  incr(ok(X)) >? ok(incr(X)) 
  s(ok(X)) >? ok(s(X)) 
  hd(ok(X)) >? ok(hd(X)) 
  tl(ok(X)) >? ok(tl(X)) 
  top(mark(X)) >? top(proper(X)) 
  top(ok(X)) >? top(active(X)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  active = \y0.y0

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