We consider the system Applicative_05__TakeDropWhile.

  Alphabet:

    cons : [a * c] --> c 
    dropWhile : [a -> b * c] --> c 
    if : [b * c * c] --> c 
    nil : [] --> c 
    takeWhile : [a -> b * c] --> c 
    true : [] --> b 

  Rules:

    if(true, x, y) => x 
    if(true, x, y) => y 
    takeWhile(f, nil) => nil 
    takeWhile(f, cons(x, y)) => if(f x, cons(x, takeWhile(f, y)), nil) 
    dropWhile(f, nil) => nil 
    dropWhile(f, cons(x, y)) => if(f x, dropWhile(f, y), cons(x, y)) 

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:

  if(true, X, Y) => X 
  if(true, X, Y) => Y 

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:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(true, %X, %Y) -> %Y
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(if(x_1, x_2, x_3)) = 2 + 2*x_1 + x_2 + x_3
 ||    POL(true) = 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:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(true, %X, %Y) -> %Y
 ||    ~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.

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] takeWhile#(F, cons(X, Y)) =#> if#(F X, cons(X, takeWhile(F, Y)), nil)   
    1] takeWhile#(F, cons(X, Y)) =#> F(X)   
    2] takeWhile#(F, cons(X, Y)) =#> takeWhile#(F, Y)   
    3] dropWhile#(F, cons(X, Y)) =#> if#(F X, dropWhile(F, Y), cons(X, Y))   
    4] dropWhile#(F, cons(X, Y)) =#> F(X)   
    5] dropWhile#(F, cons(X, Y)) =#> dropWhile#(F, Y)   

  Rules R_0:

    if(true, X, Y) => X 
    if(true, X, Y) => Y 
    takeWhile(F, nil) => nil 
    takeWhile(F, cons(X, Y)) => if(F X, cons(X, takeWhile(F, Y)), nil) 
    dropWhile(F, nil) => nil 
    dropWhile(F, cons(X, Y)) => if(F X, dropWhile(F, Y), cons(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 :  
    * 1 : 0, 1, 2, 3, 4, 5 
    * 2 : 0, 1, 2 
    * 3 :  
    * 4 : 0, 1, 2, 3, 4, 5 
    * 5 : 3, 4, 5 

This graph has the following strongly connected components:

  P_1:

    takeWhile#(F, cons(X, Y)) =#> F(X)   
    takeWhile#(F, cons(X, Y)) =#> takeWhile#(F, Y)   
    dropWhile#(F, cons(X, Y)) =#> F(X)   
    dropWhile#(F, cons(X, Y)) =#> dropWhile#(F, 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).

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 ::=

  if(true, X, Y) => X 
  if(true, X, Y) => Y 
  takeWhile(F, cons(X, Y)) => if(F X, cons(X, takeWhile(F, Y)), nil) 
  dropWhile(F, cons(X, Y)) => if(F X, dropWhile(F, Y), cons(X, 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:

  takeWhile#(F, cons(X, Y)) >? F(X) 
  takeWhile#(F, cons(X, Y)) >? takeWhile#(F, Y) 
  dropWhile#(F, cons(X, Y)) >? F(X) 
  dropWhile#(F, cons(X, Y)) >? dropWhile#(F, Y) 
  if(true, X, Y) >= X 
  if(true, X, Y) >= Y 
  takeWhile(F, cons(X, Y)) >= if(F X, cons(X, takeWhile(F, Y)), nil) 
  dropWhile(F, cons(X, Y)) >= if(F X, dropWhile(F, Y), cons(X, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + y0 + 2y1 
  dropWhile = \G0y1.2y1 
  dropWhile# = \G0y1.3 + y1 + G0(y1) 
  if = \y0y1y2.y1 + y2 
  nil = 0 
  takeWhile = \G0y1.y1 
  takeWhile# = \G0y1.3 + 2y1G0(y1) 
  true = 0 

Using this interpretation, the requirements translate to:

  [[takeWhile#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(1 + x1 + 2x2) + 2F0(1 + x1 + 2x2) + 4x2F0(1 + x1 + 2x2) > F0(x1) = [[_F0(_x1)]] 
  [[takeWhile#(_F0, cons(_x1, _x2))]] = 3 + 2x1F0(1 + x1 + 2x2) + 2F0(1 + x1 + 2x2) + 4x2F0(1 + x1 + 2x2) >= 3 + 2x2F0(x2) = [[takeWhile#(_F0, _x2)]] 
  [[dropWhile#(_F0, cons(_x1, _x2))]] = 4 + x1 + 2x2 + F0(1 + x1 + 2x2) > F0(x1) = [[_F0(_x1)]] 
  [[dropWhile#(_F0, cons(_x1, _x2))]] = 4 + x1 + 2x2 + F0(1 + x1 + 2x2) > 3 + x2 + F0(x2) = [[dropWhile#(_F0, _x2)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[takeWhile(_F0, cons(_x1, _x2))]] = 1 + x1 + 2x2 >= 1 + x1 + 2x2 = [[if(_F0 _x1, cons(_x1, takeWhile(_F0, _x2)), nil)]] 
  [[dropWhile(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 4x2 >= 1 + x1 + 4x2 = [[if(_F0 _x1, dropWhile(_F0, _x2), cons(_x1, _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_2, R_1, minimal, formative), where P_2 consists of:

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

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 apply the subterm criterion with the following projection function:

  nu(takeWhile#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(takeWhile#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(takeWhile#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_1, minimal, f) by ({}, R_1, 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.