/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> nat 
    app : [natlist * natlist] --> natlist 
    apply2 : [pair -> nat -> pair * pair * nat] --> pair 
    cons : [nat * natlist] --> natlist 
    fst : [pair] --> natlist 
    nil : [] --> natlist 
    p : [natlist * natlist] --> pair 
    pcons : [pair * plist] --> plist 
    pnil : [] --> plist 
    pps : [natlist] --> plist 
    prefixshuffle : [pair * natlist] --> plist 
    pshuffle : [natlist] --> pair 
    reverse : [natlist] --> natlist 
    s : [nat] --> nat 
    shuffle : [natlist] --> natlist 

  Rules:

    app(nil, x) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    reverse(nil) => nil 
    reverse(cons(x, y)) => app(reverse(y), cons(x, nil)) 
    shuffle(nil) => nil 
    shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) 
    fst(p(x, y)) => x 
    pshuffle(x) => p(x, shuffle(x)) 
    prefixshuffle(x, nil) => pcons(x, pnil) 
    prefixshuffle(x, cons(y, z)) => pcons(x, prefixshuffle(apply2(/\u./\v.pshuffle(app(fst(u), cons(v, nil))), x, y), reverse(z))) 
    apply2(f, x, 0) => x 
    apply2(f, x, s(y)) => f x s(y) 
    pps(x) => prefixshuffle(p(nil, nil), x) 

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:

  app(nil, X) => X 
  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
  reverse(nil) => nil 
  reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 
  shuffle(nil) => nil 
  shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
  fst(p(X, Y)) => X 
  pshuffle(X) => p(X, shuffle(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 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) QTRSRRRProof [EQUIVALENT]
 || (6) QTRS
 || (7) QTRSRRRProof [EQUIVALENT]
 || (8) QTRS
 || (9) QTRSRRRProof [EQUIVALENT]
 || (10) QTRS
 || (11) QTRSRRRProof [EQUIVALENT]
 || (12) QTRS
 || (13) QTRSRRRProof [EQUIVALENT]
 || (14) QTRS
 || (15) RisEmptyProof [EQUIVALENT]
 || (16) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    reverse(nil) -> nil
 ||    reverse(cons(%X, %Y)) -> app(reverse(%Y), cons(%X, nil))
 ||    shuffle(nil) -> nil
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 ||    fst(p(%X, %Y)) -> %X
 ||    pshuffle(%X) -> p(%X, shuffle(%X))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(app(x_1, x_2)) = x_1 + x_2
 ||    POL(cons(x_1, x_2)) = x_1 + x_2
 ||    POL(fst(x_1)) = 2 + x_1
 ||    POL(nil) = 0
 ||    POL(p(x_1, x_2)) = 2 + x_1 + x_2
 ||    POL(pshuffle(x_1)) = 2 + 2*x_1
 ||    POL(reverse(x_1)) = x_1
 ||    POL(shuffle(x_1)) = x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    fst(p(%X, %Y)) -> %X
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    reverse(nil) -> nil
 ||    reverse(cons(%X, %Y)) -> app(reverse(%Y), cons(%X, nil))
 ||    shuffle(nil) -> nil
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 ||    pshuffle(%X) -> p(%X, shuffle(%X))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(app(x_1, x_2)) = x_1 + x_2
 ||    POL(cons(x_1, x_2)) = 2*x_1 + x_2
 ||    POL(nil) = 0
 ||    POL(p(x_1, x_2)) = x_1 + x_2
 ||    POL(pshuffle(x_1)) = 1 + 2*x_1
 ||    POL(reverse(x_1)) = x_1
 ||    POL(shuffle(x_1)) = x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    pshuffle(%X) -> p(%X, shuffle(%X))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    reverse(nil) -> nil
 ||    reverse(cons(%X, %Y)) -> app(reverse(%Y), cons(%X, nil))
 ||    shuffle(nil) -> nil
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (5) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(app(x_1, x_2)) = x_1 + x_2
 ||    POL(cons(x_1, x_2)) = 2 + 2*x_1 + x_2
 ||    POL(nil) = 0
 ||    POL(reverse(x_1)) = x_1
 ||    POL(shuffle(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:
 || 
 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (6)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    reverse(nil) -> nil
 ||    reverse(cons(%X, %Y)) -> app(reverse(%Y), cons(%X, nil))
 ||    shuffle(nil) -> nil
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (7) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(app(x_1, x_2)) = x_1 + 2*x_2
 ||    POL(cons(x_1, x_2)) = x_1 + x_2
 ||    POL(nil) = 0
 ||    POL(reverse(x_1)) = 2*x_1
 ||    POL(shuffle(x_1)) = 1 + 2*x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    shuffle(nil) -> nil
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (8)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    reverse(nil) -> nil
 ||    reverse(cons(%X, %Y)) -> app(reverse(%Y), cons(%X, nil))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (9) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(app(x_1, x_2)) = x_1 + x_2
 ||    POL(cons(x_1, x_2)) = 1 + 2*x_1 + x_2
 ||    POL(nil) = 0
 ||    POL(reverse(x_1)) = 1 + x_1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    reverse(nil) -> nil
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (10)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    reverse(cons(%X, %Y)) -> app(reverse(%Y), cons(%X, nil))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (11) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(app(x_1, x_2)) = x_1 + x_2
 ||    POL(cons(x_1, x_2)) = 1 + 2*x_1 + x_2
 ||    POL(nil) = 0
 ||    POL(reverse(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:
 || 
 ||    reverse(cons(%X, %Y)) -> app(reverse(%Y), cons(%X, nil))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (12)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (13) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(app(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:
 || 
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (14)
 || Obligation:
 || Q restricted rewrite system:
 || R is empty.
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (15) RisEmptyProof (EQUIVALENT)
 || The TRS R is empty. Hence, termination is trivially proven.
 || ----------------------------------------
 || 
 || (16)
 || YES
 || 
We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]).

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

  Dependency Pairs P_0:

    0] prefixshuffle#(X, cons(Y, Z)) =#> prefixshuffle#(apply2(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y), reverse(Z))   
    1] prefixshuffle#(X, cons(Y, Z)) =#> apply2#(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y)   
    2] prefixshuffle#(X, cons(Y, Z)) =#> pshuffle#(app(fst(U), cons(V, nil)))   
    3] prefixshuffle#(X, cons(Y, Z)) =#> app#(fst(U), cons(V, nil))   
    4] prefixshuffle#(X, cons(Y, Z)) =#> fst#(U)   
    5] prefixshuffle#(X, cons(Y, Z)) =#> reverse#(Z)   
    6] pps#(X) =#> prefixshuffle#(p(nil, nil), X)   

  Rules R_0:

    app(nil, X) => X 
    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
    reverse(nil) => nil 
    reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 
    shuffle(nil) => nil 
    shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
    fst(p(X, Y)) => X 
    pshuffle(X) => p(X, shuffle(X)) 
    prefixshuffle(X, nil) => pcons(X, pnil) 
    prefixshuffle(X, cons(Y, Z)) => pcons(X, prefixshuffle(apply2(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y), reverse(Z))) 
    apply2(F, X, 0) => X 
    apply2(F, X, s(Y)) => F X s(Y) 
    pps(X) => prefixshuffle(p(nil, nil), X) 

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 
    * 1 :  
    * 2 :  
    * 3 :  
    * 4 :  
    * 5 :  
    * 6 : 0, 1, 2, 3, 4, 5 

This graph has the following strongly connected components:

  P_1:

    prefixshuffle#(X, cons(Y, Z)) =#> prefixshuffle#(apply2(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y), reverse(Z))   

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).

The formative rules of (P_1, R_0) are R_1 ::=

  app(nil, X) => X 
  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
  reverse(nil) => nil 
  reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 
  shuffle(nil) => nil 
  shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
  fst(p(X, Y)) => X 
  pshuffle(X) => p(X, shuffle(X)) 
  apply2(F, X, 0) => X 
  apply2(F, X, s(Y)) => F X s(Y) 

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].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  prefixshuffle#(X, cons(Y, Z)) >? prefixshuffle#(apply2(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y), reverse(Z)) 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  reverse(nil) >= nil 
  reverse(cons(X, Y)) >= app(reverse(Y), cons(X, nil)) 
  shuffle(nil) >= nil 
  shuffle(cons(X, Y)) >= cons(X, shuffle(reverse(Y))) 
  fst(p(X, Y)) >= X 
  pshuffle(X) >= p(X, shuffle(X)) 
  apply2(F, X, 0) >= X 
  apply2(F, X, s(Y)) >= F X s(Y) 

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

  pi( pshuffle(X) ) = #argfun-pshuffle#(p(X, shuffle(X))) 

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

The following interpretation satisfies the requirements:

  #argfun-pshuffle# = \y0.y0 
  0 = 0 
  app = \y0y1.y0 + y1 
  apply2 = \G0y1y2.y1 + G0(y1,y1) 
  cons = \y0y1.1 + y1 
  fst = \y0.y0 
  nil = 0 
  p = \y0y1.y0 
  prefixshuffle# = \y0y1.2y1 
  pshuffle = \y0.0 
  reverse = \y0.y0 
  s = \y0.0 
  shuffle = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 2 + 2x2 > 2x2 = [[prefixshuffle#(apply2(/\x./\y.#argfun-pshuffle#(p(app(fst(x), cons(y, nil)), shuffle(app(fst(x), cons(y, nil))))), _x0, _x1), reverse(_x2))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[shuffle(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 2 + 2x1 >= 1 + 2x1 = [[cons(_x0, shuffle(reverse(_x1)))]] 
  [[fst(p(_x0, _x1))]] = x0 >= x0 = [[_x0]] 
  [[#argfun-pshuffle#(p(_x0, shuffle(_x0)))]] = x0 >= x0 = [[p(_x0, shuffle(_x0))]] 
  [[apply2(_F0, _x1, 0)]] = x1 + F0(x1,x1) >= x1 = [[_x1]] 
  [[apply2(_F0, _x1, s(_x2))]] = x1 + F0(x1,x1) >= F0(x1,0) = [[_F0 _x1 s(_x2)]] 

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