/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.

  cons : [o * o] --> o 
  empty : [] --> o 
  r1 : [o * o] --> o 
  rev : [o] --> o 

  rev(X) => r1(X, empty) 
  r1(empty, X) => X 
  r1(cons(X, Y), Z) => r1(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 : [v * fa] --> fa 
  empty : [] --> fa 
  r1 : [fa * fa] --> fa 
  rev : [fa] --> fa 

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

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

  rev(X) >? r1(X, empty) 
  r1(empty, X) >? X 
  r1(cons(X, Y), Z) >? r1(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 
  empty = 0 
  r1 = \y0y1.2y1 + 3y0 
  rev = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[rev(_x0)]] = 3 + 3x0 > 3x0 = [[r1(_x0, empty)]] 
  [[r1(empty, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[r1(cons(_x0, _x1), _x2)]] = 9 + 2x2 + 3x1 + 9x0 > 6 + 2x2 + 3x1 + 6x0 = [[r1(_x1, cons(_x0, _x2))]] 

We can thus remove the following rules:

  rev(X) => r1(X, empty) 
  r1(cons(X, Y), Z) => r1(Y, cons(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]):

  r1(empty, X) >? X 

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

The following interpretation satisfies the requirements:

  empty = 3 
  r1 = \y0y1.3 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[r1(empty, _x0)]] = 6 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  r1(empty, X) => 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 +++

[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.