We consider the system Applicative_05__TreeFlatten.

  Alphabet:

    append : [a * a] --> a 
    concat : [a] --> a 
    cons : [a * a] --> a 
    flatten : [] --> a -> a 
    map : [a -> a * a] --> a 
    nil : [] --> a 
    node : [a * a] --> a 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    flatten node(x, y) => cons(x, concat(map(flatten, y))) 
    concat(nil) => nil 
    concat(cons(x, y)) => append(x, concat(y)) 
    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 

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:

  concat(nil) => nil 
  concat(cons(X, Y)) => append(X, concat(Y)) 
  append(nil, X) => X 
  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 

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:
 || 
 ||    concat(nil) -> nil
 ||    concat(cons(%X, %Y)) -> append(%X, concat(%Y))
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(append(x_1, x_2)) = 2 + 2*x_1 + x_2
 ||    POL(concat(x_1)) = 2 + 2*x_1
 ||    POL(cons(x_1, x_2)) = 1 + x_1 + x_2
 ||    POL(nil) = 1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    concat(nil) -> nil
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    concat(cons(%X, %Y)) -> append(%X, concat(%Y))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Knuth-Bendix order [KBO] with precedence:concat_1 > append_2 > cons_2
 || 
 || and weight map:
 || 
 ||    concat_1=1
 ||    cons_2=0
 ||    append_2=0
 || 
 || The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    concat(cons(%X, %Y)) -> append(%X, concat(%Y))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (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] map#(F, cons(X, Y)) =#> F(X)   
    1] map#(F, cons(X, Y)) =#> map#(F, Y)   
    2] flatten node(X, Y) =#> concat#(map(flatten, Y))   
    3] flatten node(X, Y) =#> map#(flatten, Y)   
    4] flatten node(X, Y) =#> flatten#   

  Rules R_0:

    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    flatten node(X, Y) => cons(X, concat(map(flatten, Y))) 
    concat(nil) => nil 
    concat(cons(X, Y)) => append(X, concat(Y)) 
    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 

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 
    * 1 : 0, 1 
    * 2 :  
    * 3 : 0, 1 
    * 4 :  

This graph has the following strongly connected components:

  P_1:

    map#(F, cons(X, Y)) =#> F(X)   
    map#(F, cons(X, Y)) =#> map#(F, Y)   
    flatten node(X, Y) =#> map#(flatten, 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).

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:

  map#(F, cons(X, Y)) >? F(X) 
  map#(F, cons(X, Y)) >? map#(F, Y) 
  flatten node(X, Y) >? map#(flatten, Y) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 
  flatten node(X, Y) >= cons(X, concat(map(flatten, Y))) 
  concat(nil) >= nil 
  concat(cons(X, Y)) >= append(X, concat(Y)) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 

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

The following interpretation satisfies the requirements:

  append = \y0y1.y0 + y1 
  concat = \y0.y0 
  cons = \y0y1.1 + y0 + y1 
  flatten = \y0.0 
  map = \G0y1.2 + 2y1 + G0(y1) + 3y1G0(y1) 
  map# = \G0y1.y1 + 2y1G0(y1) 
  nil = 0 
  node = \y0y1.3 + y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] 
  [[map#(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > x2 + 2x2F0(x2) = [[map#(_F0, _x2)]] 
  [[flatten node(_x0, _x1)]] = 3 + x0 + 2x1 > x1 = [[map#(flatten, _x1)]] 
  [[map(_F0, nil)]] = 2 + F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 3x1F0(1 + x1 + x2) + 3x2F0(1 + x1 + x2) + 4F0(1 + x1 + x2) >= 3 + 2x2 + F0(x2) + 3x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[flatten node(_x0, _x1)]] = 3 + x0 + 2x1 >= 3 + x0 + 2x1 = [[cons(_x0, concat(map(flatten, _x1)))]] 
  [[concat(nil)]] = 0 >= 0 = [[nil]] 
  [[concat(cons(_x0, _x1))]] = 1 + x0 + x1 >= x0 + x1 = [[append(_x0, concat(_x1))]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0).  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.