/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 
  activate : [o] --> o 
  and : [o * o] --> o 
  plus : [o * o] --> o 
  s : [o] --> o 
  tt : [] --> o 
  x : [o * o] --> o 

  and(tt, X) => activate(X) 
  plus(X, 0) => X 
  plus(X, s(Y)) => s(plus(X, Y)) 
  x(X, 0) => 0 
  x(X, s(Y)) => plus(x(X, 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 : [] --> xa 
  activate : [k] --> k 
  and : [i * k] --> k 
  plus : [xa * xa] --> xa 
  s : [xa] --> xa 
  tt : [] --> i 
  x : [xa * xa] --> xa 

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

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

  and(tt, X) >? activate(X) 
  plus(X, 0) >? X 
  plus(X, s(Y)) >? s(plus(X, Y)) 
  x(X, 0) >? 0 
  x(X, s(Y)) >? plus(x(X, Y), X) 
  activate(X) >? X 

about to try horpo

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[0]] = _|_ 
  [[activate(x_1)]] = x_1 

We choose Lex = {plus} and Mul = {and, s, tt, x}, and the following precedence: and > tt > x > plus > s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  and(tt, X) >= X 
  plus(X, _|_) >= X 
  plus(X, s(Y)) > s(plus(X, Y)) 
  x(X, _|_) >= _|_ 
  x(X, s(Y)) > plus(x(X, Y), X) 
  X >= X 

With these choices, we have:

  1] and(tt, X) >= X  because [2], by (Star) 
  2] and*(tt, X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] plus(X, _|_) >= X  because [5], by (Star) 
  5] plus*(X, _|_) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 

  7] plus(X, s(Y)) > s(plus(X, Y))  because [8], by definition 
  8] plus*(X, s(Y)) >= s(plus(X, Y))  because plus > s and [9], by (Copy) 
  9] plus*(X, s(Y)) >= plus(X, Y)  because [10], [11], [14] and [15], by (Stat) 
  10] X >= X  by (Meta) 
  11] s(Y) > Y  because [12], by definition 
  12] s*(Y) >= Y  because [13], by (Select) 
  13] Y >= Y  by (Meta) 
  14] plus*(X, s(Y)) >= X  because [10], by (Select) 
  15] plus*(X, s(Y)) >= Y  because [16], by (Select) 
  16] s(Y) >= Y  because [12], by (Star) 

  17] x(X, _|_) >= _|_  by (Bot) 

  18] x(X, s(Y)) > plus(x(X, Y), X)  because [19], by definition 
  19] x*(X, s(Y)) >= plus(x(X, Y), X)  because x > plus, [20] and [21], by (Copy) 
  20] x*(X, s(Y)) >= x(X, Y)  because x in Mul, [10] and [11], by (Stat) 
  21] x*(X, s(Y)) >= X  because [10], by (Select) 

  22] X >= X  by (Meta) 

We can thus remove the following rules:

  plus(X, s(Y)) => s(plus(X, Y)) 
  x(X, s(Y)) => plus(x(X, 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]):

  and(tt, X) >? activate(X) 
  plus(X, 0) >? X 
  x(X, 0) >? 0 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.y0 
  and = \y0y1.3 + 3y0 + 3y1 
  plus = \y0y1.3 + y0 + y1 
  tt = 3 
  x = \y0y1.3 + y0 + 3y1 

Using this interpretation, the requirements translate to:

  [[and(tt, _x0)]] = 12 + 3x0 > x0 = [[activate(_x0)]] 
  [[plus(_x0, 0)]] = 3 + x0 > x0 = [[_x0]] 
  [[x(_x0, 0)]] = 3 + x0 > 0 = [[0]] 
  [[activate(_x0)]] = x0 >= x0 = [[_x0]] 

We can thus remove the following rules:

  and(tt, X) => activate(X) 
  plus(X, 0) => X 
  x(X, 0) => 0 

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.