We consider the system iterative.

  Alphabet:

    a : [] --> o 
    b : [] --> o 
    f : [o * o -> o] --> o 
    g : [o * o * o -> o] --> o 

  Rules:

    g(x, y, h) => f(f(x, h), h) 
    f(x, h) => b 
    b => a 
    f(b, /\x.g(x, x, h)) => g(f(a, /\y.g(y, y, h)), f(b, /\z.b), h) 

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:

  b => a 

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:
 || 
 ||    b -> a
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(a) = 1
 ||    POL(b) = 2
 ||    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:
 || 
 ||    b -> a
 ||    ~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, all):

  Dependency Pairs P_0:

    0] g#(X, Y, F) =#> f#(f(X, F), F)   
    1] g#(X, Y, F) =#> f#(X, F)   
    2] f#(X, F) =#> b#   
    3] f#(b, /\x.g(x, x, F)) =#> g#(f(a, /\y.g(y, y, F)), f(b, /\z.b), F)   
    4] f#(b, /\x.g(x, x, F)) =#> f#(a, /\y.g(y, y, F))   
    5] f#(b, /\x.g(x, x, F)) =#> g#(y, y, F)   
    6] f#(b, /\x.g(x, x, F)) =#> f#(b, /\y.b)   
    7] f#(b, /\x.g(x, x, F)) =#> b#   
    8] f#(b, /\x.g(x, x, F)) =#> b#   

  Rules R_0:

    g(X, Y, F) => f(f(X, F), F) 
    f(X, F) => b 
    b => a 
    f(b, /\x.g(x, x, F)) => g(f(a, /\y.g(y, y, F)), f(b, /\z.b), F) 

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

We consider the dependency pair problem (P_0, R_0, minimal, all).

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

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

This graph has the following strongly connected components:

  P_1:

    g#(X, Y, F) =#> f#(f(X, F), F)   
    g#(X, Y, F) =#> f#(X, F)   
    f#(b, /\x.g(x, x, F)) =#> g#(f(a, /\y.g(y, y, F)), f(b, /\z.b), F)   

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, all) is finite.

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

We apply the subterm criterion with the following projection function:

  nu(f#) = 2 
  nu(g#) = 3 

Thus, we can orient the dependency pairs as follows:

  nu(g#(X, Y, F)) = F = F = nu(f#(f(X, F), F)) 
  nu(g#(X, Y, F)) = F = F = nu(f#(X, F)) 
  nu(f#(b, /\x.g(x, x, F))) = /\y.g(y, y, F) |> F = nu(g#(f(a, /\x.g(x, x, F)), f(b, /\z.b), F)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, minimal, f) by (P_2, R_0, minimal, f), where P_2 contains:

  g#(X, Y, F) =#> f#(f(X, F), F)   
  g#(X, Y, F) =#> f#(X, F)   

Thus, the original system is terminating if (P_2, R_0, minimal, all) is finite.

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

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.

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.