We consider the system 430.

  Alphabet:

    append : [natlist * natlist] --> natlist 
    cons : [nat * natlist] --> natlist 
    map : [nat -> nat * natlist] --> natlist 
    nil : [] --> natlist 

  Rules:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    append(append(X, Y), Z) => append(X, append(Y, Z)) 
    map(/\x.X(x), nil) => nil 
    map(/\x.X(x), cons(Y, Z)) => cons(X(Y), map(/\y.X(y), Z)) 

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

  append(nil, X) => X 
  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
  append(append(X, Y), Z) => append(X, append(Y, Z)) 

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:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 ||    append(append(%X, %Y), %Z) -> append(%X, append(%Y, %Z))
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Knuth-Bendix order [KBO] with precedence:~PAIR_2 > append_2 > cons_2 > nil
 || 
 || and weight map:
 || 
 ||    nil=1
 ||    append_2=0
 ||    cons_2=0
 ||    ~PAIR_2=0
 || 
 || The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 ||    append(append(%X, %Y), %Z) -> append(%X, append(%Y, %Z))
 ||    ~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] map#(/\x.X(x), cons(Y, Z)) =#> X(Y)   
    1] map#(/\x.X(x), cons(Y, Z)) =#> map#(/\y.X(y), Z)   
    2] map#(/\x.X(x), cons(Y, Z)) =#> X(y)   

  Rules R_0:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    append(append(X, Y), Z) => append(X, append(Y, Z)) 
    map(/\x.X(x), nil) => nil 
    map(/\x.X(x), cons(Y, Z)) => cons(X(Y), map(/\y.X(y), Z)) 

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 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#(/\x.X(x), cons(Y, Z)) >? X(Y) 
  map#(/\x.X(x), cons(Y, Z)) >? map#(/\y.X(y), Z) 
  map#(/\x.X(x), cons(Y, Z)) >? X(~c0) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  append(append(X, Y), Z) >= append(X, append(Y, Z)) 
  map(/\x.X(x), nil) >= nil 
  map(/\x.X(x), cons(Y, Z)) >= cons(X(Y), map(/\y.X(y), Z)) 

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

The following interpretation satisfies the requirements:

  append = \y0y1.y1 + 2y0 
  cons = \y0y1.1 + y1 + 2y0 
  map = \G0y1.3 + y1 + 2y1G0(y1) 
  map# = \G0y1.3 + G0(0) + y1G0(y1) 
  nil = 1 
  ~c0 = 0 

Using this interpretation, the requirements translate to:

  [[map#(/\x._x0(x), cons(_x1, _x2))]] = 3 + F0(0) + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) > F0(x1) = [[_x0(_x1)]] 
  [[map#(/\x._x0(x), cons(_x1, _x2))]] = 3 + F0(0) + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) >= 3 + F0(0) + x2F0(x2) = [[map#(/\x._x0(x), _x2)]] 
  [[map#(/\x._x0(x), cons(_x1, _x2))]] = 3 + F0(0) + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) > F0(0) = [[_x0(~c0)]] 
  [[append(nil, _x0)]] = 2 + x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 2 + x2 + 2x1 + 4x0 >= 1 + x2 + 2x0 + 2x1 = [[cons(_x0, append(_x1, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = x2 + 2x1 + 4x0 >= x2 + 2x0 + 2x1 = [[append(_x0, append(_x1, _x2))]] 
  [[map(/\x._x0(x), nil)]] = 4 + 2F0(1) >= 1 = [[nil]] 
  [[map(/\x._x0(x), cons(_x1, _x2))]] = 4 + x2 + 2x1 + 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 4 + x2 + 2x2F0(x2) + 2F0(x1) = [[cons(_x0(_x1), map(/\x._x0(x), _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_1, R_0, minimal, formative), where P_1 consists of:

  map#(/\x.X(x), cons(Y, Z)) =#> map#(/\y.X(y), Z)   

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).

We apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(/\x.X(x), cons(Y, Z))) = cons(Y, Z) |> Z = nu(map#(/\y.X(y), Z)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f).  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.