We consider the system Applicative_first_order_05__#3.25.

  Alphabet:

    cons : [c * d] --> d 
    f : [a] --> a 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    g : [a] --> a 
    h : [a] --> a 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    true : [] --> b 

  Rules:

    f(g(x)) => g(f(f(x))) 
    f(h(x)) => h(g(x)) 
    map(i, nil) => nil 
    map(i, cons(x, y)) => cons(i x, map(i, y)) 
    filter(i, nil) => nil 
    filter(i, cons(x, y)) => filter2(i x, i, x, y) 
    filter2(true, i, x, y) => cons(x, filter(i, y)) 
    filter2(false, i, x, y) => filter(i, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  f(g(X)) >? g(f(f(X))) 
  f(h(X)) >? h(g(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) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[f(g(_x0))]] = x0 >= x0 = [[g(f(f(_x0)))]] 
  [[f(h(_x0))]] = x0 >= x0 = [[h(g(_x0))]] 
  [[map(_F0, nil)]] = 2 + 2F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 4F0(1 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + 2x2F0(x2) + 2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 1 + 2x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2 + 2x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

  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) 

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

  f(g(X)) => g(f(f(X))) 
  f(h(X)) => h(g(X)) 

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)
 || (RULES
 || f(g(%X)) -> g(f(f(%X)))
 || f(h(%X)) -> h(g(%X))
 || )
 || 
 || Problem 1: 
 || 
 || Innermost Equivalent Processor:
 || -> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 ||  
 || 
 || Problem 1: 
 || 
 || Dependency Pairs Processor:
 || -> Pairs:
 ||  F(g(%X)) -> F(f(%X))
 ||  F(g(%X)) -> F(%X)
 || -> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || 
 || Problem 1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  F(g(%X)) -> F(f(%X))
 ||  F(g(%X)) -> F(%X)
 || -> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  F(g(%X)) -> F(f(%X))
 ||  F(g(%X)) -> F(%X)
 || ->->-> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || 
 || Problem 1: 
 || 
 || Reduction Pairs Processor:
 || -> Pairs:
 ||  F(g(%X)) -> F(f(%X))
 ||  F(g(%X)) -> F(%X)
 || -> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || -> Usable rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || ->Interpretation type:
 ||  Linear
 || ->Coefficients:
 ||  Natural Numbers
 || ->Dimension:
 ||  1
 || ->Bound:
 ||  2
 || ->Interpretation:
 ||  
 || [f](X) = X
 || [g](X) = 2.X + 2
 || [h](X) = 2
 || [F](X) = 2.X
 || 
 || Problem 1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  F(g(%X)) -> F(%X)
 || -> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  F(g(%X)) -> F(%X)
 || ->->-> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || 
 || Problem 1: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  F(g(%X)) -> F(%X)
 || -> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%X))
 || ->Projection:
 ||  pi(F) = 1
 || 
 || Problem 1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  f(g(%X)) -> g(f(f(%X)))
 ||  f(h(%X)) -> h(g(%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:


  Rules R_0:

    f(g(X)) => g(f(f(X))) 
    f(h(X)) => h(g(X)) 

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:


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 +++

[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.