/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 
  U11 : [o * o * o] --> o 
  U12 : [o * o * o] --> o 
  activate : [o] --> o 
  plus : [o * o] --> o 
  s : [o] --> o 
  tt : [] --> o 

  U11(tt, X, Y) => U12(tt, activate(X), activate(Y)) 
  U12(tt, X, Y) => s(plus(activate(Y), activate(X))) 
  plus(X, 0) => X 
  plus(X, s(Y)) => U11(tt, Y, X) 
  activate(X) => X 

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:

  0 : [] --> ab 
  U11 : [h * ab * ab] --> ab 
  U12 : [h * ab * ab] --> ab 
  activate : [ab] --> ab 
  plus : [ab * ab] --> ab 
  s : [ab] --> ab 
  tt : [] --> h 

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

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

  U11(tt, X, Y) >? U12(tt, activate(X), activate(Y)) 
  U12(tt, X, Y) >? s(plus(activate(Y), activate(X))) 
  plus(X, 0) >? X 
  plus(X, s(Y)) >? U11(tt, Y, X) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  0 = 3 
  U11 = \y0y1y2.y0 + y1 + y2 
  U12 = \y0y1y2.y0 + y1 + y2 
  activate = \y0.y0 
  plus = \y0y1.y0 + y1 
  s = \y0.y0 
  tt = 0 

Using this interpretation, the requirements translate to:

  [[U11(tt, _x0, _x1)]] = x0 + x1 >= x0 + x1 = [[U12(tt, activate(_x0), activate(_x1))]] 
  [[U12(tt, _x0, _x1)]] = x0 + x1 >= x0 + x1 = [[s(plus(activate(_x1), activate(_x0)))]] 
  [[plus(_x0, 0)]] = 3 + x0 > x0 = [[_x0]] 
  [[plus(_x0, s(_x1))]] = x0 + x1 >= x0 + x1 = [[U11(tt, _x1, _x0)]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  plus(X, 0) => 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]):

  U11(tt, X, Y) >? U12(tt, activate(X), activate(Y)) 
  U12(tt, X, Y) >? s(plus(activate(Y), activate(X))) 
  plus(X, s(Y)) >? U11(tt, Y, X) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  U11 = \y0y1y2.1 + y2 + 2y0 + 3y1 
  U12 = \y0y1y2.1 + y2 + 2y0 + 3y1 
  activate = \y0.y0 
  plus = \y0y1.1 + y0 + 3y1 
  s = \y0.1 + y0 
  tt = 1 

Using this interpretation, the requirements translate to:

  [[U11(tt, _x0, _x1)]] = 3 + x1 + 3x0 >= 3 + x1 + 3x0 = [[U12(tt, activate(_x0), activate(_x1))]] 
  [[U12(tt, _x0, _x1)]] = 3 + x1 + 3x0 > 2 + x1 + 3x0 = [[s(plus(activate(_x1), activate(_x0)))]] 
  [[plus(_x0, s(_x1))]] = 4 + x0 + 3x1 > 3 + x0 + 3x1 = [[U11(tt, _x1, _x0)]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  U12(tt, X, Y) => s(plus(activate(Y), activate(X))) 
  plus(X, s(Y)) => U11(tt, Y, 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]):

  U11(tt, X, Y) >? U12(tt, activate(X), activate(Y)) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  U11 = \y0y1y2.3 + 3y0 + 3y1 + 3y2 
  U12 = \y0y1y2.y0 + y1 + y2 
  activate = \y0.y0 
  tt = 3 

Using this interpretation, the requirements translate to:

  [[U11(tt, _x0, _x1)]] = 12 + 3x0 + 3x1 > 3 + x0 + x1 = [[U12(tt, activate(_x0), activate(_x1))]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  U11(tt, X, Y) => U12(tt, activate(X), activate(Y)) 

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

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

  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  activate = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[activate(_x0)]] = 1 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  activate(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.