/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.

  2nd : [o] --> o 
  activate : [o] --> o 
  cons : [o * o] --> o 
  from : [o] --> o 
  n!6220!6220cons : [o * o] --> o 
  n!6220!6220from : [o] --> o 
  s : [o] --> o 

  2nd(cons(X, n!6220!6220cons(Y, Z))) => activate(Y) 
  from(X) => cons(X, n!6220!6220from(s(X))) 
  cons(X, Y) => n!6220!6220cons(X, Y) 
  from(X) => n!6220!6220from(X) 
  activate(n!6220!6220cons(X, Y)) => cons(X, Y) 
  activate(n!6220!6220from(X)) => from(X) 
  activate(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]):

  2nd(cons(X, n!6220!6220cons(Y, Z))) >? activate(Y) 
  from(X) >? cons(X, n!6220!6220from(s(X))) 
  cons(X, Y) >? n!6220!6220cons(X, Y) 
  from(X) >? n!6220!6220from(X) 
  activate(n!6220!6220cons(X, Y)) >? cons(X, Y) 
  activate(n!6220!6220from(X)) >? from(X) 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  2nd = \y0.3 + 3y0 
  activate = \y0.2 + 2y0 
  cons = \y0y1.y0 + y1 
  from = \y0.2 + 2y0 
  n!6220!6220cons = \y0y1.y0 + y1 
  n!6220!6220from = \y0.y0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[2nd(cons(_x0, n!6220!6220cons(_x1, _x2)))]] = 3 + 3x0 + 3x1 + 3x2 > 2 + 2x1 = [[activate(_x1)]] 
  [[from(_x0)]] = 2 + 2x0 > 2x0 = [[cons(_x0, n!6220!6220from(s(_x0)))]] 
  [[cons(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[n!6220!6220cons(_x0, _x1)]] 
  [[from(_x0)]] = 2 + 2x0 > x0 = [[n!6220!6220from(_x0)]] 
  [[activate(n!6220!6220cons(_x0, _x1))]] = 2 + 2x0 + 2x1 > x0 + x1 = [[cons(_x0, _x1)]] 
  [[activate(n!6220!6220from(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[from(_x0)]] 
  [[activate(_x0)]] = 2 + 2x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  2nd(cons(X, n!6220!6220cons(Y, Z))) => activate(Y) 
  from(X) => cons(X, n!6220!6220from(s(X))) 
  from(X) => n!6220!6220from(X) 
  activate(n!6220!6220cons(X, Y)) => cons(X, Y) 
  activate(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]):

  cons(X, Y) >? n!6220!6220cons(X, Y) 
  activate(n!6220!6220from(X)) >? from(X) 

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

The following interpretation satisfies the requirements:

  activate = \y0.3 + y0 
  cons = \y0y1.3 + y0 + y1 
  from = \y0.y0 
  n!6220!6220cons = \y0y1.y0 + y1 
  n!6220!6220from = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[cons(_x0, _x1)]] = 3 + x0 + x1 > x0 + x1 = [[n!6220!6220cons(_x0, _x1)]] 
  [[activate(n!6220!6220from(_x0))]] = 6 + x0 > x0 = [[from(_x0)]] 

We can thus remove the following rules:

  cons(X, Y) => n!6220!6220cons(X, Y) 
  activate(n!6220!6220from(X)) => from(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 +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.