/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.

  cons : [o * o] --> o 
  empty : [] --> o 
  f : [o * o] --> o 
  g : [o * o] --> o 

  f(X, empty) => g(X, empty) 
  f(X, cons(Y, Z)) => f(cons(Y, X), Z) 
  g(empty, X) => X 
  g(cons(X, Y), Z) => g(Y, cons(X, Z)) 

As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95].  Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows:

  cons : [s * sa] --> sa 
  empty : [] --> sa 
  f : [sa * sa] --> sa 
  g : [sa * sa] --> sa 

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

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

  f(X, empty) >? g(X, empty) 
  f(X, cons(Y, Z)) >? f(cons(Y, X), Z) 
  g(empty, X) >? X 
  g(cons(X, Y), Z) >? g(Y, cons(X, Z)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.y0 + y1 
  empty = 3 
  f = \y0y1.3 + 2y0 + 2y1 
  g = \y0y1.2y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[f(_x0, empty)]] = 9 + 2x0 > 6 + 2x0 = [[g(_x0, empty)]] 
  [[f(_x0, cons(_x1, _x2))]] = 3 + 2x0 + 2x1 + 2x2 >= 3 + 2x0 + 2x1 + 2x2 = [[f(cons(_x1, _x0), _x2)]] 
  [[g(empty, _x0)]] = 6 + 2x0 > x0 = [[_x0]] 
  [[g(cons(_x0, _x1), _x2)]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[g(_x1, cons(_x0, _x2))]] 

We can thus remove the following rules:

  f(X, empty) => g(X, empty) 
  g(empty, 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]):

  f(X, cons(Y, Z)) >? f(cons(Y, X), Z) 
  g(cons(X, Y), Z) >? g(Y, cons(X, Z)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + y1 + 3y0 
  f = \y0y1.y0 + 2y1 
  g = \y0y1.2y1 + 3y0 

Using this interpretation, the requirements translate to:

  [[f(_x0, cons(_x1, _x2))]] = 6 + x0 + 2x2 + 6x1 > 3 + x0 + 2x2 + 3x1 = [[f(cons(_x1, _x0), _x2)]] 
  [[g(cons(_x0, _x1), _x2)]] = 9 + 2x2 + 3x1 + 9x0 > 6 + 2x2 + 3x1 + 6x0 = [[g(_x1, cons(_x0, _x2))]] 

We can thus remove the following rules:

  f(X, cons(Y, Z)) => f(cons(Y, X), Z) 
  g(cons(X, Y), Z) => g(Y, cons(X, 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 +++

[FuhGieParSchSwi11]  C. Fuhs, J. Giesl, M. Parting, P. Schneider-Kamp, and S. Swiderski.  Proving Termination by Dependency Pairs and Inductive Theorem Proving.  In volume 47(2) of Journal of Automated Reasoning.  133--160, 2011.
[Gra95]  B. Gramlich.  Abstract Relations Between Restricted Termination and Confluence Properties of Rewrite Systems.  In volume 24(1-2) of Fundamentae Informaticae.  3--23, 1995.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.