We consider the system Applicative_05__Ex7_9.

  Alphabet:

    0 : [] --> a 
    cons : [b * c] --> c 
    d : [a * a] --> c 
    false : [] --> c 
    filter : [b -> c * c] --> c 
    gtr : [a * a] --> c 
    if : [c * c * c] --> c 
    len : [c] --> a 
    nil : [] --> c 
    s : [a] --> a 
    sub : [a * a] --> a 
    true : [] --> c 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    sub(x, 0) => x 
    sub(s(x), s(y)) => sub(x, y) 
    gtr(0, x) => false 
    gtr(s(x), 0) => true 
    gtr(s(x), s(y)) => gtr(x, y) 
    d(x, 0) => true 
    d(s(x), s(y)) => if(gtr(x, y), false, d(s(x), sub(y, x))) 
    len(nil) => 0 
    len(cons(x, y)) => s(len(y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, 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:

  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  sub(X, 0) => X 
  sub(s(X), s(Y)) => sub(X, Y) 
  gtr(0, X) => false 
  gtr(s(X), 0) => true 
  gtr(s(X), s(Y)) => gtr(X, Y) 
  d(X, 0) => true 
  d(s(X), s(Y)) => if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
  len(nil) => 0 
  len(cons(X, Y)) => s(len(Y)) 

Moreover, the system is orthogonal.  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 terminating when seen as a many-sorted first-order TRS.

According to mutermprover, this system is indeed terminating:

 || 
 || Problem 1: 
 || 
 || (VAR %X %Y)
 || (RULES
 || d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 || d(%X,0) -> true
 || gtr(0,%X) -> false
 || gtr(s(%X),0) -> true
 || gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 || if(false,%X,%Y) -> %Y
 || if(true,%X,%Y) -> %X
 || len(cons(%X,%Y)) -> s(len(%Y))
 || len(nil) -> 0
 || sub(s(%X),s(%Y)) -> sub(%X,%Y)
 || sub(%X,0) -> %X
 || )
 || 
 || Problem 1: 
 || 
 || Innermost Equivalent Processor:
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 ||  
 || 
 || Problem 1: 
 || 
 || Dependency Pairs Processor:
 || -> Pairs:
 ||  D(s(%X),s(%Y)) -> D(s(%X),sub(%Y,%X))
 ||  D(s(%X),s(%Y)) -> GTR(%X,%Y)
 ||  D(s(%X),s(%Y)) -> IF(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  D(s(%X),s(%Y)) -> SUB(%Y,%X)
 ||  GTR(s(%X),s(%Y)) -> GTR(%X,%Y)
 ||  LEN(cons(%X,%Y)) -> LEN(%Y)
 ||  SUB(s(%X),s(%Y)) -> SUB(%X,%Y)
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || 
 || Problem 1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  D(s(%X),s(%Y)) -> D(s(%X),sub(%Y,%X))
 ||  D(s(%X),s(%Y)) -> GTR(%X,%Y)
 ||  D(s(%X),s(%Y)) -> IF(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  D(s(%X),s(%Y)) -> SUB(%Y,%X)
 ||  GTR(s(%X),s(%Y)) -> GTR(%X,%Y)
 ||  LEN(cons(%X,%Y)) -> LEN(%Y)
 ||  SUB(s(%X),s(%Y)) -> SUB(%X,%Y)
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  SUB(s(%X),s(%Y)) -> SUB(%X,%Y)
 || ->->-> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->->Cycle:
 || ->->-> Pairs:
 ||  LEN(cons(%X,%Y)) -> LEN(%Y)
 || ->->-> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->->Cycle:
 || ->->-> Pairs:
 ||  GTR(s(%X),s(%Y)) -> GTR(%X,%Y)
 || ->->-> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->->Cycle:
 || ->->-> Pairs:
 ||  D(s(%X),s(%Y)) -> D(s(%X),sub(%Y,%X))
 || ->->-> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || 
 || 
 || The problem is decomposed in 4 subproblems.
 || 
 || Problem 1.1: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  SUB(s(%X),s(%Y)) -> SUB(%X,%Y)
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Projection:
 ||  pi(SUB) = 1
 || 
 || Problem 1.1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
 || Problem 1.2: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  LEN(cons(%X,%Y)) -> LEN(%Y)
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Projection:
 ||  pi(LEN) = 1
 || 
 || Problem 1.2: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
 || Problem 1.3: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  GTR(s(%X),s(%Y)) -> GTR(%X,%Y)
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Projection:
 ||  pi(GTR) = 1
 || 
 || Problem 1.3: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
 || Problem 1.4: 
 || 
 || Reduction Pairs Processor:
 || -> Pairs:
 ||  D(s(%X),s(%Y)) -> D(s(%X),sub(%Y,%X))
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || -> Usable rules:
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Interpretation type:
 ||  Linear
 || ->Coefficients:
 ||  Natural Numbers
 || ->Dimension:
 ||  1
 || ->Bound:
 ||  2
 || ->Interpretation:
 ||  
 || [sub](X1,X2) = X1
 || [0] = 0
 || [s](X) = X + 2
 || [D](X1,X2) = 2.X2
 || 
 || Problem 1.4: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  d(s(%X),s(%Y)) -> if(gtr(%X,%Y),false,d(s(%X),sub(%Y,%X)))
 ||  d(%X,0) -> true
 ||  gtr(0,%X) -> false
 ||  gtr(s(%X),0) -> true
 ||  gtr(s(%X),s(%Y)) -> gtr(%X,%Y)
 ||  if(false,%X,%Y) -> %Y
 ||  if(true,%X,%Y) -> %X
 ||  len(cons(%X,%Y)) -> s(len(%Y))
 ||  len(nil) -> 0
 ||  sub(s(%X),s(%Y)) -> sub(%X,%Y)
 ||  sub(%X,0) -> %X
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [FuhKop19]).

We thus obtain the following dependency pair problem (P_0, R_0, computable, formative):

  Dependency Pairs P_0:

    0] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
    1] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    2] filter#(F, cons(X, Y)) =#> filter#(F, Y)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    sub(X, 0) => X 
    sub(s(X), s(Y)) => sub(X, Y) 
    gtr(0, X) => false 
    gtr(s(X), 0) => true 
    gtr(s(X), s(Y)) => gtr(X, Y) 
    d(X, 0) => true 
    d(s(X), s(Y)) => if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
    len(nil) => 0 
    len(cons(X, Y)) => s(len(Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => if(F X, cons(X, filter(F, Y)), filter(F, Y)) 

Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_0, R_0, computable, 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 
    * 2 : 0, 1, 2 

This graph has the following strongly connected components:

  P_1:

    filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    filter#(F, cons(X, Y)) =#> filter#(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, computable, formative) is finite.

We consider the dependency pair problem (P_1, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

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

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, 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.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.