/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.

  !plus : [o * o] --> o 
  0 : [] --> o 
  1 : [] --> o 
  minus : [o] --> o 

  minus(0) => 0 
  !plus(X, 0) => X 
  !plus(0, X) => X 
  !plus(minus(1), 1) => 0 
  minus(minus(X)) => X 
  !plus(X, minus(Y)) => minus(!plus(minus(X), Y)) 
  !plus(X, !plus(Y, Z)) => !plus(!plus(X, Y), Z) 
  !plus(minus(!plus(X, 1)), 1) => minus(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]):

  minus(0) >? 0 
  !plus(X, 0) >? X 
  !plus(0, X) >? X 
  !plus(minus(1), 1) >? 0 
  minus(minus(X)) >? X 
  !plus(X, minus(Y)) >? minus(!plus(minus(X), Y)) 
  !plus(X, !plus(Y, Z)) >? !plus(!plus(X, Y), Z) 
  !plus(minus(!plus(X, 1)), 1) >? minus(X) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + y1 
  0 = 0 
  1 = 3 
  minus = \y0.y0 

Using this interpretation, the requirements translate to:

  [[minus(0)]] = 0 >= 0 = [[0]] 
  [[!plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(minus(1), 1)]] = 6 > 0 = [[0]] 
  [[minus(minus(_x0))]] = x0 >= x0 = [[_x0]] 
  [[!plus(_x0, minus(_x1))]] = x0 + x1 >= x0 + x1 = [[minus(!plus(minus(_x0), _x1))]] 
  [[!plus(_x0, !plus(_x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(!plus(_x0, _x1), _x2)]] 
  [[!plus(minus(!plus(_x0, 1)), 1)]] = 6 + x0 > x0 = [[minus(_x0)]] 

We can thus remove the following rules:

  !plus(minus(1), 1) => 0 
  !plus(minus(!plus(X, 1)), 1) => minus(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]):

  minus(0) >? 0 
  !plus(X, 0) >? X 
  !plus(0, X) >? X 
  minus(minus(X)) >? X 
  !plus(X, minus(Y)) >? minus(!plus(minus(X), Y)) 
  !plus(X, !plus(Y, Z)) >? !plus(!plus(X, Y), Z) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + 2y1 
  0 = 1 
  minus = \y0.y0 

Using this interpretation, the requirements translate to:

  [[minus(0)]] = 1 >= 1 = [[0]] 
  [[!plus(_x0, 0)]] = 2 + x0 > x0 = [[_x0]] 
  [[!plus(0, _x0)]] = 1 + 2x0 > x0 = [[_x0]] 
  [[minus(minus(_x0))]] = x0 >= x0 = [[_x0]] 
  [[!plus(_x0, minus(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[minus(!plus(minus(_x0), _x1))]] 
  [[!plus(_x0, !plus(_x1, _x2))]] = x0 + 2x1 + 4x2 >= x0 + 2x1 + 2x2 = [[!plus(!plus(_x0, _x1), _x2)]] 

We can thus remove the following rules:

  !plus(X, 0) => X 
  !plus(0, X) => 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]):

  minus(0) >? 0 
  minus(minus(X)) >? X 
  !plus(X, minus(Y)) >? minus(!plus(minus(X), Y)) 
  !plus(X, !plus(Y, Z)) >? !plus(!plus(X, Y), Z) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.3 + y0 + 3y1 
  0 = 0 
  minus = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[minus(0)]] = 1 > 0 = [[0]] 
  [[minus(minus(_x0))]] = 2 + x0 > x0 = [[_x0]] 
  [[!plus(_x0, minus(_x1))]] = 6 + x0 + 3x1 > 5 + x0 + 3x1 = [[minus(!plus(minus(_x0), _x1))]] 
  [[!plus(_x0, !plus(_x1, _x2))]] = 12 + x0 + 3x1 + 9x2 > 6 + x0 + 3x1 + 3x2 = [[!plus(!plus(_x0, _x1), _x2)]] 

We can thus remove the following rules:

  minus(0) => 0 
  minus(minus(X)) => X 
  !plus(X, minus(Y)) => minus(!plus(minus(X), Y)) 
  !plus(X, !plus(Y, Z)) => !plus(!plus(X, Y), Z) 

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.