We consider the system Applicative_05__ReverseLastInit.

  Alphabet:

    compose : [a -> a * a -> a] --> a -> a 
    cons : [a * a] --> a 
    hd : [] --> a -> a 
    init : [] --> a -> a 
    last : [] --> a -> a 
    nil : [] --> a 
    reverse : [] --> a -> a 
    reverse2 : [a * a] --> a 
    tl : [] --> a -> a 

  Rules:

    compose(f, g) x => g (f x) 
    reverse x => reverse2(x, nil) 
    reverse2(nil, x) => x 
    reverse2(cons(x, y), z) => reverse2(y, cons(x, z)) 
    hd cons(x, y) => x 
    tl cons(x, y) => y 
    last => compose(hd, reverse) 
    init => compose(reverse, compose(tl, reverse)) 

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:

  reverse X => reverse2(X, nil) 
  reverse2(nil, X) => X 
  reverse2(cons(X, Y), Z) => reverse2(Y, cons(X, Z)) 
  hd cons(X, Y) => X 
  tl cons(X, Y) => 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 the external first-order termination prover, this system is indeed 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) QTRSRRRProof [EQUIVALENT]
 || (4) QTRS
 || (5) RisEmptyProof [EQUIVALENT]
 || (6) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    reverse(%X) -> reverse2(%X, nil)
 ||    reverse2(nil, %X) -> %X
 ||    reverse2(cons(%X, %Y), %Z) -> reverse2(%Y, cons(%X, %Z))
 ||    hd(cons(%X, %Y)) -> %X
 ||    tl(cons(%X, %Y)) -> %Y
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(cons(x_1, x_2)) = 1 + x_1 + x_2
 ||    POL(hd(x_1)) = 2 + x_1
 ||    POL(nil) = 2
 ||    POL(reverse(x_1)) = 2 + 2*x_1
 ||    POL(reverse2(x_1, x_2)) = 2*x_1 + x_2
 ||    POL(tl(x_1)) = 2 + x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    reverse2(nil, %X) -> %X
 ||    reverse2(cons(%X, %Y), %Z) -> reverse2(%Y, cons(%X, %Z))
 ||    hd(cons(%X, %Y)) -> %X
 ||    tl(cons(%X, %Y)) -> %Y
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    reverse(%X) -> reverse2(%X, nil)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Quasi precedence:
 || [reverse_1, nil] > reverse2_2
 || 
 || 
 || Status:
 || reverse_1: multiset status
 || reverse2_2: multiset status
 || nil: multiset status
 || 
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    reverse(%X) -> reverse2(%X, nil)
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (5) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (6)
 || 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] compose(F, G) X =#> G(F X)   
    1] compose(F, G) X =#> F(X)   
    2] last X =#> compose(hd, reverse) X   
    3] last# =#> compose#(hd, reverse)   
    4] last# =#> hd#   
    5] last# =#> reverse#   
    6] init X =#> compose(reverse, compose(tl, reverse)) X   
    7] init# =#> compose#(reverse, compose(tl, reverse))   
    8] init# =#> reverse#   
    9] init# =#> compose#(tl, reverse)   
    10] init# =#> tl#   
    11] init# =#> reverse#   

  Rules R_0:

    compose(F, G) X => G (F X) 
    reverse X => reverse2(X, nil) 
    reverse2(nil, X) => X 
    reverse2(cons(X, Y), Z) => reverse2(Y, cons(X, Z)) 
    hd cons(X, Y) => X 
    tl cons(X, Y) => Y 
    last => compose(hd, reverse) 
    init => compose(reverse, compose(tl, reverse)) 

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 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 
    * 1 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 
    * 2 : 0, 1 
    * 3 :  
    * 4 :  
    * 5 :  
    * 6 : 0, 1 
    * 7 :  
    * 8 :  
    * 9 :  
    * 10 :  
    * 11 :  

This graph has the following strongly connected components:

  P_1:

    compose(F, G) X =#> G(F X)   
    compose(F, G) X =#> F(X)   
    last X =#> compose(hd, reverse) X   
    init X =#> compose(reverse, compose(tl, reverse)) X   

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:

  compose(F, G) X >? G(F X) 
  compose(F, G) X >? F(X) 
  last(X) >? compose(hd, reverse) X 
  init(X) >? compose(reverse, compose(tl, reverse)) X 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( compose(F, G) ) = #argfun-compose#(/\x.G (F x), /\y.F y) 
  pi( init(X) ) = #argfun-init#(#argfun-compose#(/\x.#argfun-compose#(/\y.reverse (tl y), /\z.tl z) (reverse x), /\u.reverse u) X) 
  pi( last(X) ) = #argfun-last#(#argfun-compose#(/\x.reverse (hd x), /\y.hd y) X) 

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

The following interpretation satisfies the requirements:

  #argfun-compose# = \G0G1y2.3 + max(G0(y2), G1(y2)) 
  #argfun-init# = \y0.3 + y0 
  #argfun-last# = \y0.3 + y0 
  compose = \G0G1y2.0 
  hd = \y0.0 
  init = \y0.0 
  last = \y0.0 
  reverse = \y0.0 
  tl = \y0.0 

Using this interpretation, the requirements translate to:

  [[#argfun-compose#(/\x._F0 (_F1 x), /\y._F1 y) _x2]] = max(x2, 3 + max(x2, F0(max(x2, F1(x2))), F1(x2))) >= F0(max(x2, F1(x2))) = [[_F0(_F1 _x2)]] 
  [[#argfun-compose#(/\x._F0 (_F1 x), /\y._F1 y) _x2]] = max(x2, 3 + max(x2, F0(max(x2, F1(x2))), F1(x2))) >= F1(x2) = [[_F1(_x2)]] 
  [[#argfun-last#(#argfun-compose#(/\x.reverse (hd x), /\y.hd y) _x0)]] = 3 + max(x0, 3 + x0) > max(x0, 3 + x0) = [[#argfun-compose#(/\x.reverse (hd x), /\y.hd y) _x0]] 
  [[#argfun-init#(#argfun-compose#(/\x.#argfun-compose#(/\y.reverse (tl y), /\z.tl z) (reverse x), /\u.reverse u) _x0)]] = 3 + max(x0, 3 + max(x0, 3 + x0)) > max(x0, 3 + max(x0, 3 + x0)) = [[#argfun-compose#(/\x.#argfun-compose#(/\y.reverse (tl y), /\z.tl z) (reverse x), /\u.reverse u) _x0]] 

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:

  compose(F, G) X =#> F(X)   

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 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:

  compose(F, G, X) >? F(X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( compose(F, G, X) ) = #argfun-compose#(F X) 

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

The following interpretation satisfies the requirements:

  #argfun-compose# = \y0.1 + y0 
  compose = \G0G1y2.0 

Using this interpretation, the requirements translate to:

  [[#argfun-compose#(_F0 _x1)]] = 1 + max(x1, F0(x1)) > F0(x1) = [[_F0(_x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, 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.