We consider the system Applicative_first_order_05__#3.22.

  Alphabet:

    0 : [] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    plus : [a * a] --> a 
    s : [a] --> a 
    times : [a * a] --> a 
    true : [] --> b 

  Rules:

    times(x, plus(y, s(z))) => plus(times(x, plus(y, times(s(z), 0))), times(x, s(z))) 
    times(x, 0) => 0 
    times(x, s(y)) => plus(times(x, y), x) 
    plus(x, 0) => x 
    plus(x, s(y)) => s(plus(x, y)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [FuhKop19]).

We thus obtain the following dependency pair problem (P_0, R_0, computable, formative):

  Dependency Pairs P_0:

    0] times#(X, plus(Y, s(Z))) =#> plus#(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z)))   
    1] times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0)))   
    2] times#(X, plus(Y, s(Z))) =#> plus#(Y, times(s(Z), 0))   
    3] times#(X, plus(Y, s(Z))) =#> times#(s(Z), 0)   
    4] times#(X, plus(Y, s(Z))) =#> times#(X, s(Z))   
    5] times#(X, s(Y)) =#> plus#(times(X, Y), X)   
    6] times#(X, s(Y)) =#> times#(X, Y)   
    7] plus#(X, s(Y)) =#> plus#(X, Y)   
    8] map#(F, cons(X, Y)) =#> map#(F, Y)   
    9] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    10] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    11] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 
    times(X, 0) => 0 
    times(X, s(Y)) => plus(times(X, Y), X) 
    plus(X, 0) => X 
    plus(X, s(Y)) => s(plus(X, Y)) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, Y) 

Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_0, R_0, computable, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 7 
    * 1 : 0, 1, 2, 3, 4, 5, 6 
    * 2 : 7 
    * 3 :  
    * 4 : 5, 6 
    * 5 : 7 
    * 6 : 0, 1, 2, 3, 4, 5, 6 
    * 7 : 7 
    * 8 : 8 
    * 9 : 10, 11 
    * 10 : 9 
    * 11 : 9 

This graph has the following strongly connected components:

  P_1:

    times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0)))   
    times#(X, plus(Y, s(Z))) =#> times#(X, s(Z))   
    times#(X, s(Y)) =#> times#(X, Y)   

  P_2:

    plus#(X, s(Y)) =#> plus#(X, Y)   

  P_3:

    map#(F, cons(X, Y)) =#> map#(F, Y)   

  P_4:

    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(F, Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f), (P_3, R_0, m, f) and (P_4, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative) and (P_4, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_4, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_4, R_0, computable, f) by (P_5, R_0, computable, f), where P_5 contains:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative) and (P_5, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_5, R_0, computable, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

Thus, the original system is terminating if each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative) and (P_3, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_3, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, R_0, computable, f) by ({}, R_0, computable, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_2, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_2, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(plus#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(plus#(X, s(Y))) = s(Y) |> Y = nu(plus#(X, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by ({}, R_0, computable, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_1, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_1, R_0, computable, formative).

The formative rules of (P_1, R_0) are R_1 ::=

  times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 
  times(X, 0) => 0 
  times(X, s(Y)) => plus(times(X, Y), X) 
  plus(X, 0) => X 
  plus(X, s(Y)) => s(plus(X, Y)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, computable, formative) by (P_1, R_1, computable, formative).

Thus, the original system is terminating if (P_1, R_1, computable, formative) is finite.

We consider the dependency pair problem (P_1, R_1, computable, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  times#(X, plus(Y, s(Z))) >? times#(X, plus(Y, times(s(Z), 0))) 
  times#(X, plus(Y, s(Z))) >? times#(X, s(Z)) 
  times#(X, s(Y)) >? times#(X, Y) 
  times(X, plus(Y, s(Z))) >= plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 
  times(X, 0) >= 0 
  times(X, s(Y)) >= plus(times(X, Y), X) 
  plus(X, 0) >= X 
  plus(X, s(Y)) >= s(plus(X, Y)) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  plus = \y0y1.y0 + 2y1 
  s = \y0.2 + y0 
  times = \y0y1.2y0y1 
  times# = \y0y1.2y1 

Using this interpretation, the requirements translate to:

  [[times#(_x0, plus(_x1, s(_x2)))]] = 8 + 2x1 + 4x2 > 2x1 = [[times#(_x0, plus(_x1, times(s(_x2), 0)))]] 
  [[times#(_x0, plus(_x1, s(_x2)))]] = 8 + 2x1 + 4x2 > 4 + 2x2 = [[times#(_x0, s(_x2))]] 
  [[times#(_x0, s(_x1))]] = 4 + 2x1 > 2x1 = [[times#(_x0, _x1)]] 
  [[times(_x0, plus(_x1, s(_x2)))]] = 2x0x1 + 4x0x2 + 8x0 >= 2x0x1 + 4x0x2 + 8x0 = [[plus(times(_x0, plus(_x1, times(s(_x2), 0))), times(_x0, s(_x2)))]] 
  [[times(_x0, 0)]] = 0 >= 0 = [[0]] 
  [[times(_x0, s(_x1))]] = 2x0x1 + 4x0 >= 2x0 + 2x0x1 = [[plus(times(_x0, _x1), _x0)]] 
  [[plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[plus(_x0, s(_x1))]] = 4 + x0 + 2x1 >= 2 + x0 + 2x1 = [[s(plus(_x0, _x1))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.