TRS Stand 20472 pair #381712623

details

property	value
status	complete
benchmark	log2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.555384159088 seconds
cpu usage	0.542761823
max memory	1.5597568E7

stage attributes

key	value
output-size	23030
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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.

  !940 : [] --> o 
  !minus : [o * o] --> o 
  !plus : [o * o] --> o 
  0 : [o] --> o 
  1 : [o] --> o 
  false : [] --> o 
  ge : [o * o] --> o 
  if : [o * o * o] --> o 
  log : [o] --> o 
  log!450 : [o] --> o 
  not : [o] --> o 
  true : [] --> o 

  0(!940) => !940 
  !plus(!940, X) => X 
  !plus(X, !940) => X 
  !plus(0(X), 0(Y)) => 0(!plus(X, Y)) 
  !plus(0(X), 1(Y)) => 1(!plus(X, Y)) 
  !plus(1(X), 0(Y)) => 1(!plus(X, Y)) 
  !plus(1(X), 1(Y)) => 0(!plus(!plus(X, Y), 1(!940))) 
  !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) 
  !minus(!940, X) => !940 
  !minus(X, !940) => X 
  !minus(0(X), 0(Y)) => 0(!minus(X, Y)) 
  !minus(0(X), 1(Y)) => 1(!minus(!minus(X, Y), 1(!940))) 
  !minus(1(X), 0(Y)) => 1(!minus(X, Y)) 
  !minus(1(X), 1(Y)) => 0(!minus(X, Y)) 
  not(true) => false 
  not(false) => true 
  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  ge(0(X), 0(Y)) => ge(X, Y) 
  ge(0(X), 1(Y)) => not(ge(Y, X)) 
  ge(1(X), 0(Y)) => ge(X, Y) 
  ge(1(X), 1(Y)) => ge(X, Y) 
  ge(X, !940) => true 
  ge(!940, 0(X)) => ge(!940, X) 
  ge(!940, 1(X)) => false 
  log(X) => !minus(log!450(X), 1(!940)) 
  log!450(!940) => !940 
  log!450(1(X)) => !plus(log!450(X), 1(!940)) 
  log!450(0(X)) => if(ge(X, 1(!940)), !plus(log!450(X), 1(!940)), !940) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  0(!940) >? !940 
  !plus(!940, X) >? X 
  !plus(X, !940) >? X 
  !plus(0(X), 0(Y)) >? 0(!plus(X, Y)) 
  !plus(0(X), 1(Y)) >? 1(!plus(X, Y)) 
  !plus(1(X), 0(Y)) >? 1(!plus(X, Y)) 
  !plus(1(X), 1(Y)) >? 0(!plus(!plus(X, Y), 1(!940))) 
  !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) 
  !minus(!940, X) >? !940 
  !minus(X, !940) >? X 
  !minus(0(X), 0(Y)) >? 0(!minus(X, Y)) 
  !minus(0(X), 1(Y)) >? 1(!minus(!minus(X, Y), 1(!940))) 
  !minus(1(X), 0(Y)) >? 1(!minus(X, Y)) 
  !minus(1(X), 1(Y)) >? 0(!minus(X, Y)) 
  not(true) >? false 
  not(false) >? true 
  if(true, X, Y) >? X 
  if(false, X, Y) >? Y 
  ge(0(X), 0(Y)) >? ge(X, Y) 
  ge(0(X), 1(Y)) >? not(ge(Y, X)) 
  ge(1(X), 0(Y)) >? ge(X, Y) 
  ge(1(X), 1(Y)) >? ge(X, Y) 
  ge(X, !940) >? true 
  ge(!940, 0(X)) >? ge(!940, X) 
  ge(!940, 1(X)) >? false 
  log(X) >? !minus(log!450(X), 1(!940)) 
  log!450(!940) >? !940 
  log!450(1(X)) >? !plus(log!450(X), 1(!940)) 
  log!450(0(X)) >? if(ge(X, 1(!940)), !plus(log!450(X), 1(!940)), !940) 

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

The following interpretation satisfies the requirements:

  !940 = 0 
  !minus = \y0y1.y0 + y1 
  !plus = \y0y1.y0 + y1 
  0 = \y0.2y0 
  1 = \y0.2y0 
  false = 0 
  ge = \y0y1.y0 + y1 
  if = \y0y1y2.y0 + y1 + y2

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