/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.

  !dot : [o * o] --> o 
  !plus!plus : [o * o] --> o 
  car : [o] --> o 
  cdr : [o] --> o 
  false : [] --> o 
  nil : [] --> o 
  null : [o] --> o 
  rev : [o] --> o 
  true : [] --> o 

  rev(nil) => nil 
  rev(!dot(X, Y)) => !plus!plus(rev(Y), !dot(X, nil)) 
  car(!dot(X, Y)) => X 
  cdr(!dot(X, Y)) => Y 
  null(nil) => true 
  null(!dot(X, Y)) => false 
  !plus!plus(nil, X) => X 
  !plus!plus(!dot(X, Y), Z) => !dot(X, !plus!plus(Y, 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:

  !dot : [q * tb] --> tb 
  !plus!plus : [tb * tb] --> tb 
  car : [tb] --> q 
  cdr : [tb] --> tb 
  false : [] --> cb 
  nil : [] --> tb 
  null : [tb] --> cb 
  rev : [tb] --> tb 
  true : [] --> cb 

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(nil) >? nil 
  rev(!dot(X, Y)) >? !plus!plus(rev(Y), !dot(X, nil)) 
  car(!dot(X, Y)) >? X 
  cdr(!dot(X, Y)) >? Y 
  null(nil) >? true 
  null(!dot(X, Y)) >? false 
  !plus!plus(nil, X) >? X 
  !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.y0 + y1 
  !plus!plus = \y0y1.y0 + 2y1 
  car = \y0.3 + 2y0 
  cdr = \y0.3 + 2y0 
  false = 0 
  nil = 0 
  null = \y0.3 + 3y0 
  rev = \y0.2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[rev(nil)]] = 0 >= 0 = [[nil]] 
  [[rev(!dot(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!plus!plus(rev(_x1), !dot(_x0, nil))]] 
  [[car(!dot(_x0, _x1))]] = 3 + 2x0 + 2x1 > x0 = [[_x0]] 
  [[cdr(!dot(_x0, _x1))]] = 3 + 2x0 + 2x1 > x1 = [[_x1]] 
  [[null(nil)]] = 3 > 0 = [[true]] 
  [[null(!dot(_x0, _x1))]] = 3 + 3x0 + 3x1 > 0 = [[false]] 
  [[!plus!plus(nil, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[!plus!plus(!dot(_x0, _x1), _x2)]] = x0 + x1 + 2x2 >= x0 + x1 + 2x2 = [[!dot(_x0, !plus!plus(_x1, _x2))]] 

We can thus remove the following rules:

  car(!dot(X, Y)) => X 
  cdr(!dot(X, Y)) => Y 
  null(nil) => true 
  null(!dot(X, Y)) => false 

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(nil) >? nil 
  rev(!dot(X, Y)) >? !plus!plus(rev(Y), !dot(X, nil)) 
  !plus!plus(nil, X) >? X 
  !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.y0 + y1 
  !plus!plus = \y0y1.y0 + y1 
  nil = 0 
  rev = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[rev(nil)]] = 1 > 0 = [[nil]] 
  [[rev(!dot(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[!plus!plus(rev(_x1), !dot(_x0, nil))]] 
  [[!plus!plus(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[!plus!plus(!dot(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!dot(_x0, !plus!plus(_x1, _x2))]] 

We can thus remove the following rules:

  rev(nil) => nil 

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(!dot(X, Y)) >? !plus!plus(rev(Y), !dot(X, nil)) 
  !plus!plus(nil, X) >? X 
  !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.1 + y1 + 2y0 
  !plus!plus = \y0y1.y0 + y1 
  nil = 0 
  rev = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[rev(!dot(_x0, _x1))]] = 2 + 2x1 + 4x0 > 1 + 2x0 + 2x1 = [[!plus!plus(rev(_x1), !dot(_x0, nil))]] 
  [[!plus!plus(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[!plus!plus(!dot(_x0, _x1), _x2)]] = 1 + x1 + x2 + 2x0 >= 1 + x1 + x2 + 2x0 = [[!dot(_x0, !plus!plus(_x1, _x2))]] 

We can thus remove the following rules:

  rev(!dot(X, Y)) => !plus!plus(rev(Y), !dot(X, nil)) 

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

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

  !plus!plus(nil, X) >? X 
  !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) 

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

The following interpretation satisfies the requirements:

  !dot = \y0y1.3 + y0 + y1 
  !plus!plus = \y0y1.3 + y1 + 3y0 
  nil = 3 

Using this interpretation, the requirements translate to:

  [[!plus!plus(nil, _x0)]] = 12 + x0 > x0 = [[_x0]] 
  [[!plus!plus(!dot(_x0, _x1), _x2)]] = 12 + x2 + 3x0 + 3x1 > 6 + x0 + x2 + 3x1 = [[!dot(_x0, !plus!plus(_x1, _x2))]] 

We can thus remove the following rules:

  !plus!plus(nil, X) => X 
  !plus!plus(!dot(X, Y), Z) => !dot(X, !plus!plus(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 +++

[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.