We consider the system AotoYamada_05__012.

  Alphabet:

    and : [c * c] --> c 
    cons : [a * b] --> b 
    false : [] --> c 
    forall : [a -> c * b] --> c 
    forsome : [a -> c * b] --> c 
    nil : [] --> b 
    or : [c * c] --> c 
    true : [] --> c 

  Rules:

    and(true, true) => true 
    and(x, false) => false 
    and(false, x) => false 
    or(true, x) => true 
    or(x, true) => true 
    or(false, false) => false 
    forall(f, nil) => true 
    forall(f, cons(x, y)) => and(f x, forall(f, y)) 
    forsome(f, nil) => false 
    forsome(f, cons(x, y)) => or(f x, forsome(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:

  and(true, true) => true 
  and(X, false) => false 
  and(false, X) => false 
  or(true, X) => true 
  or(X, true) => true 
  or(false, false) => false 

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:
 || 
 ||    and(true, true) -> true
 ||    and(%X, false) -> false
 ||    and(false, %X) -> false
 ||    or(true, %X) -> true
 ||    or(%X, true) -> true
 ||    or(false, false) -> false
 ||    ~PAIR(%X, %Y) -> %X
 ||    ~PAIR(%X, %Y) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(and(x_1, x_2)) = 1 + 2*x_1 + x_2
 ||    POL(false) = 2
 ||    POL(or(x_1, x_2)) = 1 + 2*x_1 + 2*x_2
 ||    POL(true) = 0
 ||    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:
 || 
 ||    and(true, true) -> true
 ||    and(%X, false) -> false
 ||    and(false, %X) -> false
 ||    or(true, %X) -> true
 ||    or(%X, true) -> true
 ||    or(false, false) -> false
 ||    ~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] forall#(F, cons(X, Y)) =#> and#(F X, forall(F, Y))   
    1] forall#(F, cons(X, Y)) =#> F(X)   
    2] forall#(F, cons(X, Y)) =#> forall#(F, Y)   
    3] forsome#(F, cons(X, Y)) =#> or#(F X, forsome(F, Y))   
    4] forsome#(F, cons(X, Y)) =#> F(X)   
    5] forsome#(F, cons(X, Y)) =#> forsome#(F, Y)   

  Rules R_0:

    and(true, true) => true 
    and(X, false) => false 
    and(false, X) => false 
    or(true, X) => true 
    or(X, true) => true 
    or(false, false) => false 
    forall(F, nil) => true 
    forall(F, cons(X, Y)) => and(F X, forall(F, Y)) 
    forsome(F, nil) => false 
    forsome(F, cons(X, Y)) => or(F X, forsome(F, 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:

    forall#(F, cons(X, Y)) =#> F(X)   
    forall#(F, cons(X, Y)) =#> forall#(F, Y)   
    forsome#(F, cons(X, Y)) =#> F(X)   
    forsome#(F, cons(X, Y)) =#> forsome#(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).

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:

  forall#(F, cons(X, Y)) >? F(X) 
  forall#(F, cons(X, Y)) >? forall#(F, Y) 
  forsome#(F, cons(X, Y)) >? F(X) 
  forsome#(F, cons(X, Y)) >? forsome#(F, Y) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + y0 + 2y1 
  forall# = \G0y1.3 + 2y1 + y1G0(y1) 
  forsome# = \G0y1.3 + y1 + 2y1G0(y1) 

Using this interpretation, the requirements translate to:

  [[forall#(_F0, cons(_x1, _x2))]] = 9 + 2x1 + 4x2 + 2x2F0(3 + x1 + 2x2) + 3F0(3 + x1 + 2x2) + x1F0(3 + x1 + 2x2) > F0(x1) = [[_F0(_x1)]] 
  [[forall#(_F0, cons(_x1, _x2))]] = 9 + 2x1 + 4x2 + 2x2F0(3 + x1 + 2x2) + 3F0(3 + x1 + 2x2) + x1F0(3 + x1 + 2x2) > 3 + 2x2 + x2F0(x2) = [[forall#(_F0, _x2)]] 
  [[forsome#(_F0, cons(_x1, _x2))]] = 6 + x1 + 2x2 + 2x1F0(3 + x1 + 2x2) + 4x2F0(3 + x1 + 2x2) + 6F0(3 + x1 + 2x2) > F0(x1) = [[_F0(_x1)]] 
  [[forsome#(_F0, cons(_x1, _x2))]] = 6 + x1 + 2x2 + 2x1F0(3 + x1 + 2x2) + 4x2F0(3 + x1 + 2x2) + 6F0(3 + x1 + 2x2) > 3 + x2 + 2x2F0(x2) = [[forsome#(_F0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, 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.