We consider the system Applicative_first_order_05__02.

  Alphabet:

    !facdot : [a * a] --> a 
    1 : [] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    i : [a] --> a 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    true : [] --> b 

  Rules:

    !facdot(1, x) => x 
    !facdot(x, 1) => x 
    !facdot(i(x), x) => 1 
    !facdot(x, i(x)) => 1 
    !facdot(i(x), !facdot(x, y)) => y 
    !facdot(x, !facdot(i(x), y)) => y 
    i(1) => 1 
    i(i(x)) => x 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, 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:

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

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, all):

  Dependency Pairs P_0:

    0] map#(F, cons(X, Y)) =#> F(X)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   
    2] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    3] filter#(F, cons(X, Y)) =#> F(X)   
    4] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    5] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    !facdot(1, X) => X 
    !facdot(X, 1) => X 
    !facdot(i(X), X) => 1 
    !facdot(X, i(X)) => 1 
    !facdot(i(X), !facdot(X, Y)) => Y 
    !facdot(X, !facdot(i(X), Y)) => Y 
    i(1) => 1 
    i(i(X)) => X 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, Y) 

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 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:

  map#(F, cons(X, Y)) >? F(X) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) 
  filter#(F, cons(X, Y)) >? F(X) 
  filter2#(true, F, X, Y) >? filter#(F, Y) 
  filter2#(false, F, X, Y) >? filter#(F, Y) 
  !facdot(1, X) >= X 
  !facdot(X, 1) >= X 
  !facdot(i(X), X) >= 1 
  !facdot(X, i(X)) >= 1 
  !facdot(i(X), !facdot(X, Y)) >= Y 
  !facdot(X, !facdot(i(X), Y)) >= Y 
  i(1) >= 1 
  i(i(X)) >= X 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >= cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >= filter(F, Y) 
  filter(F, X) >= filter#(F, X) 
  filter2(X, F, Y, Z) >= filter2#(X, F, Y, Z) 
  map(F, X) >= map#(F, X) 

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

The following interpretation satisfies the requirements:

  !facdot = \y0y1.3 + 3y0 + 3y1 
  1 = 0 
  cons = \y0y1.3 + y0 + y1 
  false = 3 
  filter = \G0y1.2y1 + G0(0) + 3y1G0(y1) 
  filter2 = \y0G1y2y3.y0 + y2 + 2y3 + G1(0) + 3y3G1(y3) 
  filter2# = \y0G1y2y3.y3 + y3G1(y3) 
  filter# = \G0y1.y1 + y1G0(y1) 
  i = \y0.3 + 3y0 
  map = \G0y1.2 + 3y1 + G0(0) + G0(y1) + y1G0(y1) 
  map# = \G0y1.y1 + G0(y1) 
  nil = 0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + F0(3 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + F0(3 + x1 + x2) > x2 + F0(x2) = [[map#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + 3F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) > x2 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + 3F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[filter2#(true, _F0, _x1, _x2)]] = x2 + x2F0(x2) >= x2 + x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[filter2#(false, _F0, _x1, _x2)]] = x2 + x2F0(x2) >= x2 + x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[!facdot(1, _x0)]] = 3 + 3x0 >= x0 = [[_x0]] 
  [[!facdot(_x0, 1)]] = 3 + 3x0 >= x0 = [[_x0]] 
  [[!facdot(i(_x0), _x0)]] = 12 + 12x0 >= 0 = [[1]] 
  [[!facdot(_x0, i(_x0))]] = 12 + 12x0 >= 0 = [[1]] 
  [[!facdot(i(_x0), !facdot(_x0, _x1))]] = 21 + 9x1 + 18x0 >= x1 = [[_x1]] 
  [[!facdot(_x0, !facdot(i(_x0), _x1))]] = 39 + 9x1 + 30x0 >= x1 = [[_x1]] 
  [[i(1)]] = 3 >= 0 = [[1]] 
  [[i(i(_x0))]] = 12 + 9x0 >= x0 = [[_x0]] 
  [[map(_F0, nil)]] = 2 + 2F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 11 + 3x1 + 3x2 + F0(0) + 4F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) >= 5 + 3x2 + F0(0) + F0(x2) + x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 9F0(3 + x1 + x2) >= x1 + 2x2 + F0(0) + 3x2F0(x2) + max(x1, F0(x1)) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 3 + x1 + 2x2 + F0(0) + 3x2F0(x2) >= 3 + x1 + 2x2 + F0(0) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 3 + x1 + 2x2 + F0(0) + 3x2F0(x2) >= 2x2 + F0(0) + 3x2F0(x2) = [[filter(_F0, _x2)]] 
  [[filter(_F0, _x1)]] = 2x1 + F0(0) + 3x1F0(x1) >= x1 + x1F0(x1) = [[filter#(_F0, _x1)]] 
  [[filter2(_x0, _F1, _x2, _x3)]] = x0 + x2 + 2x3 + F1(0) + 3x3F1(x3) >= x3 + x3F1(x3) = [[filter2#(_x0, _F1, _x2, _x3)]] 
  [[map(_F0, _x1)]] = 2 + 3x1 + F0(0) + F0(x1) + x1F0(x1) >= x1 + F0(x1) = [[map#(_F0, _x1)]] 

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, all) by (P_1, R_0, minimal, all), where P_1 consists of:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

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 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 +++

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