TRS Stand 20472 pair #381714245

details

property	value
status	complete
benchmark	2.35.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n024.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0368530750275 seconds
cpu usage	0.029825933
max memory	1400832.0

stage attributes

key	value
output-size	2624
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.

  and : [o * o] --> o 
  false : [] --> o 
  implies : [o * o] --> o 
  not : [o] --> o 

  and(X, false) => false 
  and(X, not(false)) => X 
  not(not(X)) => X 
  implies(false, X) => not(false) 
  implies(X, false) => not(X) 
  implies(not(X), not(Y)) => implies(Y, and(X, Y)) 

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

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

  and(X, false) >? false 
  and(X, not(false)) >? X 
  not(not(X)) >? X 
  implies(false, X) >? not(false) 
  implies(X, false) >? not(X) 
  implies(not(X), not(Y)) >? implies(Y, and(X, Y)) 

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

The following interpretation satisfies the requirements:

  and = \y0y1.y0 + y1 
  false = 0 
  implies = \y0y1.3 + 2y0 + 2y1 
  not = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[and(_x0, false)]] = x0 >= 0 = [[false]] 
  [[and(_x0, not(false))]] = 1 + x0 > x0 = [[_x0]] 
  [[not(not(_x0))]] = 3 + 4x0 > x0 = [[_x0]] 
  [[implies(false, _x0)]] = 3 + 2x0 > 1 = [[not(false)]] 
  [[implies(_x0, false)]] = 3 + 2x0 > 1 + 2x0 = [[not(_x0)]] 
  [[implies(not(_x0), not(_x1))]] = 7 + 4x0 + 4x1 > 3 + 2x0 + 4x1 = [[implies(_x1, and(_x0, _x1))]] 

We can thus remove the following rules:

  and(X, not(false)) => X 
  not(not(X)) => X 
  implies(false, X) => not(false) 
  implies(X, false) => not(X) 
  implies(not(X), not(Y)) => implies(Y, and(X, Y)) 

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

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

  and(X, false) >? false 

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

The following interpretation satisfies the requirements:

  and = \y0y1.3 + y0 + 3y1 
  false = 0 

Using this interpretation, the requirements translate to:

  [[and(_x0, false)]] = 3 + x0 > 0 = [[false]] 

We can thus remove the following rules:

  and(X, false) => false 

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.

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