We consider the system AotoYamada_05__013.

  Alphabet:

    append : [c * c] --> c 
    cons : [b * c] --> c 
    flatwith : [a -> b * b] --> c 
    flatwithsub : [a -> b * c] --> c 
    leaf : [a] --> b 
    nil : [] --> c 
    node : [c] --> b 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    flatwith(f, leaf(x)) => cons(f x, nil) 
    flatwith(f, node(x)) => flatwithsub(f, x) 
    flatwithsub(f, nil) => nil 
    flatwithsub(f, cons(x, y)) => append(flatwith(f, x), flatwithsub(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:

  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) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    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(cons(x_1, x_2)) = 1 + 2*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:
 || 
 ||    append(nil, %X) -> %X
 ||    append(cons(%X, %Y), %Z) -> cons(%X, append(%Y, %Z))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (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] flatwith#(F, leaf(X)) =#> F(X)   
    1] flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
    2] flatwithsub#(F, cons(X, Y)) =#> append#(flatwith(F, X), flatwithsub(F, Y))   
    3] flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
    4] flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

  Rules R_0:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    flatwith(F, leaf(X)) => cons(F X, nil) 
    flatwith(F, node(X)) => flatwithsub(F, X) 
    flatwithsub(F, nil) => nil 
    flatwithsub(F, cons(X, Y)) => append(flatwith(F, X), flatwithsub(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 : 0, 1, 2, 3, 4 
    * 1 : 2, 3, 4 
    * 2 :  
    * 3 : 0, 1 
    * 4 : 2, 3, 4 

This graph has the following strongly connected components:

  P_1:

    flatwith#(F, leaf(X)) =#> F(X)   
    flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
    flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
    flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(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).

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:

  flatwith#(F, leaf(X)) >? F(X) 
  flatwith#(F, node(X)) >? flatwithsub#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? flatwith#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? flatwithsub#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  flatwith(F, leaf(X)) >= cons(F X, nil) 
  flatwith(F, node(X)) >= flatwithsub(F, X) 
  flatwithsub(F, nil) >= nil 
  flatwithsub(F, cons(X, Y)) >= append(flatwith(F, X), flatwithsub(F, Y)) 

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

The following interpretation satisfies the requirements:

  append = \y0y1.y0 + y1 
  cons = \y0y1.1 + y1 + 2y0 
  flatwith = \G0y1.3y1 + 2y1G0(y1) 
  flatwith# = \G0y1.1 + G0(y1) + 2y1G0(y1) 
  flatwithsub = \G0y1.2y1 + 2y1G0(y1) 
  flatwithsub# = \G0y1.1 + 2y1G0(y1) 
  leaf = \y0.3 + 3y0 
  nil = 1 
  node = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[flatwith#(_F0, leaf(_x1))]] = 1 + 6x1F0(3 + 3x1) + 7F0(3 + 3x1) > F0(x1) = [[_F0(_x1)]] 
  [[flatwith#(_F0, node(_x1))]] = 1 + 2x1F0(3 + x1) + 7F0(3 + x1) >= 1 + 2x1F0(x1) = [[flatwithsub#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 1 + 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 1 + F0(x1) + 2x1F0(x1) = [[flatwith#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 1 + 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 1 + 2x2F0(x2) = [[flatwithsub#(_F0, _x2)]] 
  [[append(nil, _x0)]] = 1 + x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 + 2x0 >= 1 + x1 + x2 + 2x0 = [[cons(_x0, append(_x1, _x2))]] 
  [[flatwith(_F0, leaf(_x1))]] = 9 + 9x1 + 6x1F0(3 + 3x1) + 6F0(3 + 3x1) >= 2 + 2max(x1, F0(x1)) = [[cons(_F0 _x1, nil)]] 
  [[flatwith(_F0, node(_x1))]] = 9 + 3x1 + 2x1F0(3 + x1) + 6F0(3 + x1) >= 2x1 + 2x1F0(x1) = [[flatwithsub(_F0, _x1)]] 
  [[flatwithsub(_F0, nil)]] = 2 + 2F0(1) >= 1 = [[nil]] 
  [[flatwithsub(_F0, cons(_x1, _x2))]] = 2 + 2x2 + 4x1 + 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 2x2 + 3x1 + 2x1F0(x1) + 2x2F0(x2) = [[append(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_2, R_0, minimal, formative), where P_2 consists of:

  flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
  flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
  flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

Thus, the original system is terminating if (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(flatwith#) = 2 
  nu(flatwithsub#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(flatwith#(F, node(X))) = node(X) |> X = nu(flatwithsub#(F, X)) 
  nu(flatwithsub#(F, cons(X, Y))) = cons(X, Y) |> X = nu(flatwith#(F, X)) 
  nu(flatwithsub#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(flatwithsub#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, 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.