/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.

  !plus : [o * o] --> o 
  0 : [] --> o 
  bin : [o * o] --> o 
  s : [o] --> o 

  bin(X, 0) => s(0) 
  bin(0, s(X)) => 0 
  bin(s(X), s(Y)) => !plus(bin(X, s(Y)), bin(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]):

  bin(X, 0) >? s(0) 
  bin(0, s(X)) >? 0 
  bin(s(X), s(Y)) >? !plus(bin(X, s(Y)), bin(X, Y)) 

about to try horpo

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {!plus, 0, bin, s}, and the following precedence: bin > !plus > 0 > s

With these choices, we have:

  1] bin(X, 0) > s(0)  because [2], by definition 
  2] bin*(X, 0) >= s(0)  because bin > s and [3], by (Copy) 
  3] bin*(X, 0) >= 0  because bin > 0, by (Copy) 

  4] bin(0, s(X)) >= 0  because [5], by (Star) 
  5] bin*(0, s(X)) >= 0  because bin > 0, by (Copy) 

  6] bin(s(X), s(Y)) >= !plus(bin(X, s(Y)), bin(X, Y))  because [7], by (Star) 
  7] bin*(s(X), s(Y)) >= !plus(bin(X, s(Y)), bin(X, Y))  because bin > !plus, [8] and [14], by (Copy) 
  8] bin*(s(X), s(Y)) >= bin(X, s(Y))  because bin in Mul, [9] and [12], by (Stat) 
  9] s(X) > X  because [10], by definition 
  10] s*(X) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 
  12] s(Y) >= s(Y)  because s in Mul and [13], by (Fun) 
  13] Y >= Y  by (Meta) 
  14] bin*(s(X), s(Y)) >= bin(X, Y)  because bin in Mul, [15] and [16], by (Stat) 
  15] s(X) >= X  because [10], by (Star) 
  16] s(Y) > Y  because [17], by definition 
  17] s*(Y) >= Y  because [13], by (Select) 

We can thus remove the following rules:

  bin(X, 0) => s(0) 

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

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

  bin(0, s(X)) >? 0 
  bin(s(X), s(Y)) >? !plus(bin(X, s(Y)), bin(X, Y)) 

about to try horpo

We use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {} and Mul = {!plus, 0, bin, s}, and the following precedence: bin > !plus > 0 > s

With these choices, we have:

  1] bin(0, s(X)) >= 0  because [2], by (Star) 
  2] bin*(0, s(X)) >= 0  because bin > 0, by (Copy) 

  3] bin(s(X), s(Y)) > !plus(bin(X, s(Y)), bin(X, Y))  because [4], by definition 
  4] bin*(s(X), s(Y)) >= !plus(bin(X, s(Y)), bin(X, Y))  because bin > !plus, [5] and [11], by (Copy) 
  5] bin*(s(X), s(Y)) >= bin(X, s(Y))  because bin in Mul, [6] and [9], by (Stat) 
  6] s(X) > X  because [7], by definition 
  7] s*(X) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 
  9] s(Y) >= s(Y)  because s in Mul and [10], by (Fun) 
  10] Y >= Y  by (Meta) 
  11] bin*(s(X), s(Y)) >= bin(X, Y)  because bin in Mul, [6] and [12], by (Stat) 
  12] s(Y) >= Y  because [13], by (Star) 
  13] s*(Y) >= Y  because [10], by (Select) 

We can thus remove the following rules:

  bin(s(X), s(Y)) => !plus(bin(X, s(Y)), bin(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]):

  bin(0, s(X)) >? 0 

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

The following interpretation satisfies the requirements:

  0 = 0 
  bin = \y0y1.3 + 3y0 + 3y1 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[bin(0, s(_x0))]] = 12 + 3x0 > 0 = [[0]] 

We can thus remove the following rules:

  bin(0, s(X)) => 0 

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.