We consider the system onearg.

  Alphabet:

    0 : [] --> nat 
    add : [nat] --> nat -> nat 
    eq : [nat] --> nat -> bool 
    err : [] --> nat 
    false : [] --> bool 
    id : [] --> nat -> nat 
    nul : [] --> nat -> bool 
    pred : [nat] --> nat 
    s : [nat] --> nat 
    true : [] --> bool 

  Rules:

    nul 0 => true 
    nul s(x) => false 
    nul err => false 
    pred(0) => err 
    pred(s(x)) => x 
    id x => x 
    eq(0) => nul 
    eq(s(x)) => /\y.eq(x) pred(y) 
    add(0) => id 
    add(s(x)) => /\y.add(x) s(y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 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]):

  nul 0 >? true 
  nul s(X) >? false 
  nul err >? false 
  pred(0) >? err 
  pred(s(X)) >? X 
  id X >? X 
  eq(0) >? nul 
  eq(s(X)) >? /\x.eq(X) pred(x) 
  add(0) >? id 
  add(s(X)) >? /\x.add(X) s(x) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  add = \y0y1.3 + y1 + 3y0 + 3y0y1 
  eq = \y0y1.3 + y0 + 2y0y1 
  err = 0 
  false = 0 
  id = \y0.0 
  nul = \y0.0 
  pred = \y0.2y0 
  s = \y0.1 + 2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[nul 0]] = 3 > 0 = [[true]] 
  [[nul s(_x0)]] = 1 + 2x0 > 0 = [[false]] 
  [[nul err]] = 0 >= 0 = [[false]] 
  [[pred(0)]] = 6 > 0 = [[err]] 
  [[pred(s(_x0))]] = 2 + 4x0 > x0 = [[_x0]] 
  [[id _x0]] = x0 >= x0 = [[_x0]] 
  [[eq(0)]] = \y0.6 + 6y0 > \y0.0 = [[nul]] 
  [[eq(s(_x0))]] = \y0.4 + 2y0 + 2x0 + 4y0x0 > \y0.3 + x0 + 2y0 + 4x0y0 = [[/\x.eq(_x0) pred(x)]] 
  [[add(0)]] = \y0.12 + 10y0 > \y0.0 = [[id]] 
  [[add(s(_x0))]] = \y0.6 + 4y0 + 6y0x0 + 6x0 > \y0.5 + 4y0 + 6x0 + 6x0y0 = [[/\x.add(_x0) s(x)]] 

We can thus remove the following rules:

  nul 0 => true 
  nul s(X) => false 
  pred(0) => err 
  pred(s(X)) => X 
  eq(0) => nul 
  eq(s(X)) => /\x.eq(X) pred(x) 
  add(0) => id 
  add(s(X)) => /\x.add(X) s(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]):

  nul(err) >? false 
  id(X) >? X 

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

The following interpretation satisfies the requirements:

  err = 3 
  false = 0 
  id = \y0.3 + y0 
  nul = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[nul(err)]] = 12 > 0 = [[false]] 
  [[id(_x0)]] = 3 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  nul(err) => false 
  id(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 +++

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