We consider the system average.

  Alphabet:

    0 : [] --> nat 
    apply : [nat * nat] --> nat 
    avg : [nat * nat] --> nat 
    check : [nat] --> nat 
    fun : [nat -> nat] --> nat 
    s : [nat] --> nat 

  Rules:

    avg(s(x), y) => avg(x, s(y)) 
    avg(x, s(s(s(y)))) => s(avg(s(x), y)) 
    avg(0, 0) => 0 
    avg(0, s(0)) => 0 
    avg(0, s(s(0))) => s(0) 
    apply(fun(f), x) => f check(x) 
    check(s(x)) => s(check(x)) 
    check(0) => 0 

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

We observe that the rules contain a first-order subset:

  avg(s(X), Y) => avg(X, s(Y)) 
  avg(X, s(s(s(Y)))) => s(avg(s(X), Y)) 
  avg(0, 0) => 0 
  avg(0, s(0)) => 0 
  avg(0, s(s(0))) => s(0) 
  check(s(X)) => s(check(X)) 
  check(0) => 0 

Moreover, the system is finitely branching.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is Ce-terminating when seen as a many-sorted first-order TRS.

According to mutermprover, this system is indeed Ce-terminating:

 || 
 || Problem 1: 
 || 
 || (VAR %X %Y)
 || (RULES
 || avg(0,0) -> 0
 || avg(0,s(0)) -> 0
 || avg(0,s(s(0))) -> s(0)
 || avg(s(%X),%Y) -> avg(%X,s(%Y))
 || avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 || check(0) -> 0
 || check(s(%X)) -> s(check(%X))
 || ~PAIR(%X,%Y) -> %X
 || ~PAIR(%X,%Y) -> %Y
 || )
 || 
 || Problem 1: 
 || 
 || Dependency Pairs Processor:
 || -> Pairs:
 ||  AVG(s(%X),%Y) -> AVG(%X,s(%Y))
 ||  AVG(%X,s(s(s(%Y)))) -> AVG(s(%X),%Y)
 ||  CHECK(s(%X)) -> CHECK(%X)
 || -> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || 
 || Problem 1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  AVG(s(%X),%Y) -> AVG(%X,s(%Y))
 ||  AVG(%X,s(s(s(%Y)))) -> AVG(s(%X),%Y)
 ||  CHECK(s(%X)) -> CHECK(%X)
 || -> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  CHECK(s(%X)) -> CHECK(%X)
 || ->->-> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->->Cycle:
 || ->->-> Pairs:
 ||  AVG(s(%X),%Y) -> AVG(%X,s(%Y))
 ||  AVG(%X,s(s(s(%Y)))) -> AVG(s(%X),%Y)
 || ->->-> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || 
 || 
 || The problem is decomposed in 2 subproblems.
 || 
 || Problem 1.1: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  CHECK(s(%X)) -> CHECK(%X)
 || -> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Projection:
 ||  pi(CHECK) = 1
 || 
 || Problem 1.1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
 || Problem 1.2: 
 || 
 || Reduction Pair Processor:
 || -> Pairs:
 ||  AVG(s(%X),%Y) -> AVG(%X,s(%Y))
 ||  AVG(%X,s(s(s(%Y)))) -> AVG(s(%X),%Y)
 || -> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || -> Usable rules:
 ||  Empty
 || ->Interpretation type:
 ||  Linear
 || ->Coefficients:
 ||  Natural Numbers
 || ->Dimension:
 ||  1
 || ->Bound:
 ||  2
 || ->Interpretation:
 ||  
 || [s](X) = X + 2
 || [AVG](X1,X2) = 2.X1 + X2
 || 
 || Problem 1.2: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  AVG(%X,s(s(s(%Y)))) -> AVG(s(%X),%Y)
 || -> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  AVG(%X,s(s(s(%Y)))) -> AVG(s(%X),%Y)
 || ->->-> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || 
 || Problem 1.2: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  AVG(%X,s(s(s(%Y)))) -> AVG(s(%X),%Y)
 || -> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Projection:
 ||  pi(AVG) = 2
 || 
 || Problem 1.2: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  avg(0,0) -> 0
 ||  avg(0,s(0)) -> 0
 ||  avg(0,s(s(0))) -> s(0)
 ||  avg(s(%X),%Y) -> avg(%X,s(%Y))
 ||  avg(%X,s(s(s(%Y)))) -> s(avg(s(%X),%Y))
 ||  check(0) -> 0
 ||  check(s(%X)) -> s(check(%X))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
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] apply#(fun(F), X) =#> F(check(X))   
    1] apply#(fun(F), X) =#> check#(X)   

  Rules R_0:

    avg(s(X), Y) => avg(X, s(Y)) 
    avg(X, s(s(s(Y)))) => s(avg(s(X), Y)) 
    avg(0, 0) => 0 
    avg(0, s(0)) => 0 
    avg(0, s(s(0))) => s(0) 
    apply(fun(F), X) => F check(X) 
    check(s(X)) => s(check(X)) 
    check(0) => 0 

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 
    * 1 :  

This graph has the following strongly connected components:

  P_1:

    apply#(fun(F), X) =#> F(check(X))   

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).

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 ::=

  apply(fun(F), X) => F check(X) 

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:

  apply#(fun(F), X) >? F(check(X)) 
  apply(fun(F), X) >= F check(X) 

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

The following interpretation satisfies the requirements:

  apply = \y0y1.3 + 3y0 
  apply# = \y0y1.3 + y0 
  check = \y0.0 
  fun = \G0.3 + G0(0) 

Using this interpretation, the requirements translate to:

  [[apply#(fun(_F0), _x1)]] = 6 + F0(0) > F0(0) = [[_F0(check(_x1))]] 
  [[apply(fun(_F0), _x1)]] = 12 + 3F0(0) >= F0(0) = [[_F0 check(_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 +++

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