/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 
  cons : [o] --> o 
  filter : [o * o * o] --> o 
  nats : [o] --> o 
  s : [o] --> o 
  sieve : [o] --> o 
  zprimes : [] --> o 

  filter(cons(X), 0, Y) => cons(0) 
  filter(cons(X), s(Y), Z) => cons(X) 
  sieve(cons(0)) => cons(0) 
  sieve(cons(s(X))) => cons(s(X)) 
  nats(X) => cons(X) 
  zprimes => sieve(nats(s(s(0)))) 

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 : [] --> db 
  cons : [db] --> fb 
  filter : [fb * db * o] --> fb 
  nats : [db] --> fb 
  s : [db] --> db 
  sieve : [fb] --> fb 
  zprimes : [] --> fb 

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

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

  filter(cons(X), 0, Y) >? cons(0) 
  filter(cons(X), s(Y), Z) >? cons(X) 
  sieve(cons(0)) >? cons(0) 
  sieve(cons(s(X))) >? cons(s(X)) 
  nats(X) >? cons(X) 
  zprimes >? sieve(nats(s(s(0)))) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  cons = \y0.y0 
  filter = \y0y1y2.3 + y2 + 3y0 + 3y1 
  nats = \y0.2y0 
  s = \y0.y0 
  sieve = \y0.y0 
  zprimes = 3 

Using this interpretation, the requirements translate to:

  [[filter(cons(_x0), 0, _x1)]] = 3 + x1 + 3x0 > 0 = [[cons(0)]] 
  [[filter(cons(_x0), s(_x1), _x2)]] = 3 + x2 + 3x0 + 3x1 > x0 = [[cons(_x0)]] 
  [[sieve(cons(0))]] = 0 >= 0 = [[cons(0)]] 
  [[sieve(cons(s(_x0)))]] = x0 >= x0 = [[cons(s(_x0))]] 
  [[nats(_x0)]] = 2x0 >= x0 = [[cons(_x0)]] 
  [[zprimes]] = 3 > 0 = [[sieve(nats(s(s(0))))]] 

We can thus remove the following rules:

  filter(cons(X), 0, Y) => cons(0) 
  filter(cons(X), s(Y), Z) => cons(X) 
  zprimes => sieve(nats(s(s(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]):

  sieve(cons(0)) >? cons(0) 
  sieve(cons(s(X))) >? cons(s(X)) 
  nats(X) >? cons(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  cons = \y0.y0 
  nats = \y0.3 + 3y0 
  s = \y0.y0 
  sieve = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[sieve(cons(0))]] = 3 > 0 = [[cons(0)]] 
  [[sieve(cons(s(_x0)))]] = 3 + 3x0 > x0 = [[cons(s(_x0))]] 
  [[nats(_x0)]] = 3 + 3x0 > x0 = [[cons(_x0)]] 

We can thus remove the following rules:

  sieve(cons(0)) => cons(0) 
  sieve(cons(s(X))) => cons(s(X)) 
  nats(X) => cons(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.