We consider the system 454.

  Alphabet:

    a : [] --> o 
    f : [o] --> o 
    g : [o -> o] --> o 
    h : [o] --> o 
    hprime : [o] --> o 

  Rules:

    f(g(/\x.X(x))) => f(X(X(a))) 
    g(/\x.hprime(x)) => h(h(a)) 
    h(X) => hprime(X) 

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

  h(X) => hprime(X) 

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 the external first-order termination prover, this system is indeed Ce-terminating:

 || proof of resources/system.trs
 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
 || 
 || 
 || Termination w.r.t. Q of the given QTRS could be proven:
 || 
 || (0) QTRS
 || (1) QTRSRRRProof [EQUIVALENT]
 || (2) QTRS
 || (3) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    h(%X) -> hprime(%X)
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(h(x_1)) = 2 + 2*x_1
 ||    POL(hprime(x_1)) = 1 + x_1
 ||    POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    h(%X) -> hprime(%X)
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (4)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

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

  Dependency Pairs P_0:

    0] f#(g(/\x.X(x))) =#> f#(X(X(a)))   
    1] f#(g(/\x.X(x))) =#> X(X(a))   
    2] f#(g(/\x.X(x))) =#> X(a) {X : 1}  
    3] g#(/\x.hprime(x)) =#> h#(h(a))   
    4] g#(/\x.hprime(x)) =#> h#(a)   

  Rules R_0:

    f(g(/\x.X(x))) => f(X(X(a))) 
    g(/\x.hprime(x)) => h(h(a)) 
    h(X) => hprime(X) 

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 
    * 1 : 0, 1, 2, 3, 4 
    * 2 : 0, 1, 2, 3, 4 
    * 3 :  
    * 4 :  

This graph has the following strongly connected components:

  P_1:

    f#(g(/\x.X(x))) =#> f#(X(X(a)))   
    f#(g(/\x.X(x))) =#> X(X(a))   
    f#(g(/\x.X(x))) =#> X(a) {X : 1}  

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).

This combination (P_1, R_0) has no formative rules!  We will name the empty set of rules:R_1.

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:

  f#(g(/\x.X(x))) >? f#(X(X(a))) 
  f#(g(/\x.X(x))) >? X(X(a)) 
  f#(g(/\x.X(x))) >? X(a) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[a]] = _|_ 

We choose Lex = {} and Mul = {f#, g}, and the following precedence: f# > g

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  f#(g(/\x.X(x))) >= f#(X(X(_|_))) 
  f#(g(/\x.X(x))) >= X(X(_|_)) 
  f#(g(/\x.X(x))) > X(_|_) 

With these choices, we have:

  1] f#(g(/\x.X(x))) >= f#(X(X(_|_)))  because f# in Mul and [2], by (Fun) 
  2] g(/\x.X(x)) >= X(X(_|_))  because [3], by (Star) 
  3] g*(/\x.X(x)) >= X(X(_|_))  because [4], by (Select) 
  4] X(g*(/\x.X(x))) >= X(X(_|_))  because [5], by (Meta) 
  5] g*(/\x.X(x)) >= X(_|_)  because [6], by (Select) 
  6] X(g*(/\x.X(x))) >= X(_|_)  because [7], by (Meta) 
  7] g*(/\x.X(x)) >= _|_  by (Bot) 

  8] f#(g(/\x.X(x))) >= X(X(_|_))  because [9], by (Star) 
  9] f#*(g(/\x.X(x))) >= X(X(_|_))  because [2], by (Select) 

  10] f#(g(/\x.X(x))) > X(_|_)  because [11], by definition 
  11] f#*(g(/\x.X(x))) >= X(_|_)  because [12], by (Select) 
  12] g(/\x.X(x)) >= X(_|_)  because [5], by (Star) 

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_2, R_1, minimal, formative), where P_2 consists of:

  f#(g(/\x.X(x))) =#> f#(X(X(a)))   
  f#(g(/\x.X(x))) =#> X(X(a))   

Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_2, 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:

  f#(g(/\x.X(x))) >? f#(X(X(a))) 
  f#(g(/\x.X(x))) >? X(X(a)) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[a]] = _|_ 

We choose Lex = {} and Mul = {f#, g}, and the following precedence: f# > g

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  f#(g(/\x.X(x))) >= f#(X(X(_|_))) 
  f#(g(/\x.X(x))) > X(X(_|_)) 

With these choices, we have:

  1] f#(g(/\x.X(x))) >= f#(X(X(_|_)))  because f# in Mul and [2], by (Fun) 
  2] g(/\x.X(x)) >= X(X(_|_))  because [3], by (Star) 
  3] g*(/\x.X(x)) >= X(X(_|_))  because [4], by (Select) 
  4] X(g*(/\x.X(x))) >= X(X(_|_))  because [5], by (Meta) 
  5] g*(/\x.X(x)) >= X(_|_)  because [6], by (Select) 
  6] X(g*(/\x.X(x))) >= X(_|_)  because [7], by (Meta) 
  7] g*(/\x.X(x)) >= _|_  by (Bot) 

  8] f#(g(/\x.X(x))) > X(X(_|_))  because [9], by definition 
  9] f#*(g(/\x.X(x))) >= X(X(_|_))  because [2], by (Select) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_1, minimal, formative) by (P_3, R_1, minimal, formative), where P_3 consists of:

  f#(g(/\x.X(x))) =#> f#(X(X(a)))   

Thus, the original system is terminating if (P_3, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_1, minimal, 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:

  f#(g(/\x.X(x))) >? f#(X(X(a))) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[a]] = _|_ 

We choose Lex = {} and Mul = {f#, g}, and the following precedence: g > f#

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  f#(g(/\x.X(x))) > f#(X(X(_|_))) 

With these choices, we have:

  1] f#(g(/\x.X(x))) > f#(X(X(_|_)))  because [2], by definition 
  2] f#*(g(/\x.X(x))) >= f#(X(X(_|_)))  because [3], by (Select) 
  3] g(/\x.X(x)) >= f#(X(X(_|_)))  because [4], by (Star) 
  4] g*(/\x.X(x)) >= f#(X(X(_|_)))  because g > f# and [5], by (Copy) 
  5] g*(/\x.X(x)) >= X(X(_|_))  because [6], by (Select) 
  6] X(g*(/\x.X(x))) >= X(X(_|_))  because [7], by (Meta) 
  7] g*(/\x.X(x)) >= X(_|_)  because [8], by (Select) 
  8] X(g*(/\x.X(x))) >= X(_|_)  because [9], by (Meta) 
  9] g*(/\x.X(x)) >= _|_  by (Bot) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, 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.