We consider the system Applicative_05__TreeSize.

  Alphabet:

    0 : [] --> c 
    cons : [c * b] --> b 
    map : [c -> c * b] --> b 
    nil : [] --> b 
    node : [a * b] --> c 
    plus : [c * c] --> c 
    s : [c] --> c 
    size : [] --> c -> c 
    sum : [b] --> c 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    sum(cons(x, y)) => plus(x, sum(y)) 
    size node(x, y) => s(sum(map(size, y))) 
    plus(0, x) => 0 
    plus(s(x), y) => s(plus(x, 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 dynamic dependency pairs.

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] map#(F, cons(X, Y)) =#> F(X)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   
    2] sum#(cons(X, Y)) =#> plus#(X, sum(Y))   
    3] sum#(cons(X, Y)) =#> sum#(Y)   
    4] size node(X, Y) =#> sum#(map(size, Y))   
    5] size node(X, Y) =#> map#(size, Y)   
    6] size node(X, Y) =#> size#   
    7] plus#(s(X), Y) =#> plus#(X, Y)   

  Rules R_0:

    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    sum(cons(X, Y)) => plus(X, sum(Y)) 
    size node(X, Y) => s(sum(map(size, Y))) 
    plus(0, X) => 0 
    plus(s(X), Y) => s(plus(X, Y)) 

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

We consider the dependency pair problem (P_0, R_0, minimal, 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 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 1 : 0, 1 
    * 2 : 7 
    * 3 : 2, 3 
    * 4 : 2, 3 
    * 5 : 0, 1 
    * 6 :  
    * 7 : 7 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    sum#(cons(X, Y)) =#> sum#(Y)   

  P_3:

    plus#(s(X), Y) =#> plus#(X, 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) and (P_3, R_0, m, f).

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

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

We apply the subterm criterion with the following projection function:

  nu(plus#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by ({}, R_0, minimal, 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, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(sum#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(sum#(cons(X, Y))) = cons(X, Y) |> Y = nu(sum#(Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, 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, minimal, formative) is finite.

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

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

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

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

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

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  map#(F, cons(X, Y)) >? F(X) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  size node(X, Y) >? map#(size, Y) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.2 + y0 + 2y1 
  map = \G0y1.y1 + y1G0(y1) 
  map# = \G0y1.2G0(y1) 
  node = \y0y1.3 
  size = \y0.0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 2F0(2 + x1 + 2x2) >= F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 2F0(2 + x1 + 2x2) >= 2F0(x2) = [[map#(_F0, _x2)]] 
  [[size node(_x0, _x1)]] = 3 > 0 = [[map#(size, _x1)]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + 2x2 + 2x2F0(2 + x1 + 2x2) + 2F0(2 + x1 + 2x2) + x1F0(2 + x1 + 2x2) >= 2 + 2x2 + 2x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 consists of:

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

Thus, the original system is terminating if (P_4, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

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

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

The following interpretation satisfies the requirements:

  cons = \y0y1.2 + y1 + 2y0 
  map = \G0y1.y1 + 2y1G0(y1) 
  map# = \G0y1.3 + y1 + y1G0(y1) 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > 3 + x2 + x2F0(x2) = [[map#(_F0, _x2)]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) >= 2 + x2 + 2x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, 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 +++

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