/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.

  0 : [] --> o 
  f : [o * o] --> o 
  g : [o * o] --> o 
  nil : [] --> o 
  norm : [o] --> o 
  rem : [o * o] --> o 
  s : [o] --> o 

  norm(nil) => 0 
  norm(g(X, Y)) => s(norm(X)) 
  f(X, nil) => g(nil, X) 
  f(X, g(Y, Z)) => g(f(X, Y), Z) 
  rem(nil, X) => nil 
  rem(g(X, Y), 0) => g(X, Y) 
  rem(g(X, Y), s(Z)) => rem(X, Z) 

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

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

  norm(nil) >? 0 
  norm(g(X, Y)) >? s(norm(X)) 
  f(X, nil) >? g(nil, X) 
  f(X, g(Y, Z)) >? g(f(X, Y), Z) 
  rem(nil, X) >? nil 
  rem(g(X, Y), 0) >? g(X, Y) 
  rem(g(X, Y), s(Z)) >? rem(X, Z) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  f = \y0y1.2 + 2y0 + 3y1 
  g = \y0y1.2 + y0 + 2y1 
  nil = 0 
  norm = \y0.y0 
  rem = \y0y1.3 + y1 + 2y0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[norm(nil)]] = 0 >= 0 = [[0]] 
  [[norm(g(_x0, _x1))]] = 2 + x0 + 2x1 > 1 + x0 = [[s(norm(_x0))]] 
  [[f(_x0, nil)]] = 2 + 2x0 >= 2 + 2x0 = [[g(nil, _x0)]] 
  [[f(_x0, g(_x1, _x2))]] = 8 + 2x0 + 3x1 + 6x2 > 4 + 2x0 + 2x2 + 3x1 = [[g(f(_x0, _x1), _x2)]] 
  [[rem(nil, _x0)]] = 3 + x0 > 0 = [[nil]] 
  [[rem(g(_x0, _x1), 0)]] = 7 + 2x0 + 4x1 > 2 + x0 + 2x1 = [[g(_x0, _x1)]] 
  [[rem(g(_x0, _x1), s(_x2))]] = 8 + x2 + 2x0 + 4x1 > 3 + x2 + 2x0 = [[rem(_x0, _x2)]] 

We can thus remove the following rules:

  norm(g(X, Y)) => s(norm(X)) 
  f(X, g(Y, Z)) => g(f(X, Y), Z) 
  rem(nil, X) => nil 
  rem(g(X, Y), 0) => g(X, Y) 
  rem(g(X, Y), s(Z)) => rem(X, Z) 

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

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

  norm(nil) >? 0 
  f(X, nil) >? g(nil, X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  f = \y0y1.3 + y0 + 3y1 
  g = \y0y1.y0 + y1 
  nil = 0 
  norm = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[norm(nil)]] = 3 > 0 = [[0]] 
  [[f(_x0, nil)]] = 3 + x0 > x0 = [[g(nil, _x0)]] 

We can thus remove the following rules:

  norm(nil) => 0 
  f(X, nil) => g(nil, X) 

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.