We consider the system prefixshuffle.

  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 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 [Kop13]).

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

  Dependency Pairs P_0:

    0] app#(cons(X, Y), Z) =#> app#(Y, Z)   
    1] reverse#(cons(X, Y)) =#> app#(reverse(Y), cons(X, nil))   
    2] reverse#(cons(X, Y)) =#> reverse#(Y)   
    3] shuffle#(cons(X, Y)) =#> shuffle#(reverse(Y))   
    4] shuffle#(cons(X, Y)) =#> reverse#(Y)   
    5] pshuffle#(X) =#> shuffle#(X)   
    6] prefixshuffle#(X, cons(Y, Z)) =#> prefixshuffle#(apply2(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y), reverse(Z))   
    7] prefixshuffle#(X, cons(Y, Z)) =#> apply2#(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y)   
    8] prefixshuffle#(X, cons(Y, Z)) =#> pshuffle#(app(fst(U), cons(V, nil)))   
    9] prefixshuffle#(X, cons(Y, Z)) =#> app#(fst(U), cons(V, nil))   
    10] prefixshuffle#(X, cons(Y, Z)) =#> fst#(U)   
    11] prefixshuffle#(X, cons(Y, Z)) =#> reverse#(Z)   
    12] 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, static, formative) is finite.

We consider the dependency pair problem (P_0, R_0, static, formative).

The formative rules of (P_0, 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_0, R_0, static, formative) by (P_0, R_1, static, formative).

Thus, the original system is terminating if (P_0, R_1, static, formative) is finite.

We consider the dependency pair problem (P_0, R_1, static, 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:

  app#(cons(X, Y), Z) >? app#(Y, Z) 
  reverse#(cons(X, Y)) >? app#(reverse(Y), cons(X, nil)) 
  reverse#(cons(X, Y)) >? reverse#(Y) 
  shuffle#(cons(X, Y)) >? shuffle#(reverse(Y)) 
  shuffle#(cons(X, Y)) >? reverse#(Y) 
  pshuffle#(X) >? shuffle#(X) 
  prefixshuffle#(X, cons(Y, Z)) >? prefixshuffle#(apply2(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y), reverse(Z)) 
  prefixshuffle#(X, cons(Y, Z)) >? apply2#(/\x./\y.pshuffle(app(fst(x), cons(y, nil))), X, Y) 
  prefixshuffle#(X, cons(Y, Z)) >? pshuffle#(app(fst(U), cons(V, nil))) 
  prefixshuffle#(X, cons(Y, Z)) >? app#(fst(U), cons(V, nil)) 
  prefixshuffle#(X, cons(Y, Z)) >? fst#(U) 
  prefixshuffle#(X, cons(Y, Z)) >? reverse#(Z) 
  pps#(X) >? prefixshuffle#(p(nil, nil), X) 
  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( pps#(X) ) = #argfun-pps##(prefixshuffle#(p(nil, nil), X)) 
  pi( pshuffle(X) ) = #argfun-pshuffle#(p(X, shuffle(X))) 
  pi( pshuffle#(X) ) = #argfun-pshuffle##(shuffle#(X)) 

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

The following interpretation satisfies the requirements:

  #argfun-pps## = \y0.3 + y0 
  #argfun-pshuffle# = \y0.y0 
  #argfun-pshuffle## = \y0.y0 
  0 = 0 
  app = \y0y1.2 + y1 
  app# = \y0y1.0 
  apply2 = \G0y1y2.y1 + 2G0(y1,y2) 
  apply2# = \G0y1y2.0 
  cons = \y0y1.1 
  fst = \y0.y0 
  fst# = \y0.0 
  nil = 0 
  p = \y0y1.y0 
  pps# = \y0.0 
  prefixshuffle# = \y0y1.3 
  pshuffle = \y0.0 
  pshuffle# = \y0.0 
  reverse = \y0.3 + y0 
  reverse# = \y0.0 
  s = \y0.0 
  shuffle = \y0.1 
  shuffle# = \y0.2 

Using this interpretation, the requirements translate to:

  [[app#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[app#(_x1, _x2)]] 
  [[reverse#(cons(_x0, _x1))]] = 0 >= 0 = [[app#(reverse(_x1), cons(_x0, nil))]] 
  [[reverse#(cons(_x0, _x1))]] = 0 >= 0 = [[reverse#(_x1)]] 
  [[shuffle#(cons(_x0, _x1))]] = 2 >= 2 = [[shuffle#(reverse(_x1))]] 
  [[shuffle#(cons(_x0, _x1))]] = 2 > 0 = [[reverse#(_x1)]] 
  [[#argfun-pshuffle##(shuffle#(_x0))]] = 2 >= 2 = [[shuffle#(_x0)]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 3 >= 3 = [[prefixshuffle#(apply2(/\x./\y.#argfun-pshuffle#(p(app(fst(x), cons(y, nil)), shuffle(app(fst(x), cons(y, nil))))), _x0, _x1), reverse(_x2))]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 3 > 0 = [[apply2#(/\x./\y.#argfun-pshuffle#(p(app(fst(x), cons(y, nil)), shuffle(app(fst(x), cons(y, nil))))), _x0, _x1)]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 3 > 2 = [[#argfun-pshuffle##(shuffle#(app(fst(_x3), cons(_x4, nil))))]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 3 > 0 = [[app#(fst(_x3), cons(_x4, nil))]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 3 > 0 = [[fst#(_x3)]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 3 > 0 = [[reverse#(_x2)]] 
  [[#argfun-pps##(prefixshuffle#(p(nil, nil), _x0))]] = 6 > 3 = [[prefixshuffle#(p(nil, nil), _x0)]] 
  [[app(nil, _x0)]] = 2 + x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 2 + x2 >= 1 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 3 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 4 >= 3 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[shuffle(nil)]] = 1 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 1 >= 1 = [[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 + 2F0(x1,0) >= x1 = [[_x1]] 
  [[apply2(_F0, _x1, s(_x2))]] = x1 + 2F0(x1,0) >= F0(x1,0) = [[_F0 _x1 s(_x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, static, formative) by (P_1, R_1, static, formative), where P_1 consists of:

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

Thus, the original system is terminating if (P_1, R_1, static, formative) is finite.

We consider the dependency pair problem (P_1, R_1, static, 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:

  app#(cons(X, Y), Z) >? app#(Y, Z) 
  reverse#(cons(X, Y)) >? app#(reverse(Y), cons(X, nil)) 
  reverse#(cons(X, Y)) >? reverse#(Y) 
  shuffle#(cons(X, Y)) >? shuffle#(reverse(Y)) 
  pshuffle#(X) >? shuffle#(X) 
  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))) 
  pi( pshuffle#(X) ) = #argfun-pshuffle##(shuffle#(X)) 

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

The following interpretation satisfies the requirements:

  #argfun-pshuffle# = \y0.y0 
  #argfun-pshuffle## = \y0.3 + y0 
  0 = 0 
  app = \y0y1.y1 
  app# = \y0y1.0 
  apply2 = \G0y1y2.y1 + G0(y1,y1) 
  cons = \y0y1.0 
  fst = \y0.y0 
  nil = 0 
  p = \y0y1.y0 
  prefixshuffle# = \y0y1.0 
  pshuffle = \y0.0 
  pshuffle# = \y0.0 
  reverse = \y0.2 
  reverse# = \y0.3 
  s = \y0.0 
  shuffle = \y0.2 
  shuffle# = \y0.0 

Using this interpretation, the requirements translate to:

  [[app#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[app#(_x1, _x2)]] 
  [[reverse#(cons(_x0, _x1))]] = 3 > 0 = [[app#(reverse(_x1), cons(_x0, nil))]] 
  [[reverse#(cons(_x0, _x1))]] = 3 >= 3 = [[reverse#(_x1)]] 
  [[shuffle#(cons(_x0, _x1))]] = 0 >= 0 = [[shuffle#(reverse(_x1))]] 
  [[#argfun-pshuffle##(shuffle#(_x0))]] = 3 > 0 = [[shuffle#(_x0)]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[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)]] = x2 >= 0 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 2 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 2 >= 0 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[shuffle(nil)]] = 2 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 2 >= 0 = [[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 the dependency pair problem (P_1, R_1, static, formative) by (P_2, R_1, static, formative), where P_2 consists of:

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

Thus, the original system is terminating if (P_2, R_1, static, formative) is finite.

We consider the dependency pair problem (P_2, R_1, static, 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:

  app#(cons(X, Y), Z) >? app#(Y, Z) 
  reverse#(cons(X, Y)) >? reverse#(Y) 
  shuffle#(cons(X, Y)) >? shuffle#(reverse(Y)) 
  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 
  app# = \y0y1.y0 
  apply2 = \G0y1y2.y1 + 2G0(y1,y1) 
  cons = \y0y1.2 + y1 
  fst = \y0.y0 
  nil = 0 
  p = \y0y1.y0 
  prefixshuffle# = \y0y1.0 
  pshuffle = \y0.0 
  reverse = \y0.y0 
  reverse# = \y0.0 
  s = \y0.0 
  shuffle = \y0.y0 
  shuffle# = \y0.0 

Using this interpretation, the requirements translate to:

  [[app#(cons(_x0, _x1), _x2)]] = 2 + x1 > x1 = [[app#(_x1, _x2)]] 
  [[reverse#(cons(_x0, _x1))]] = 0 >= 0 = [[reverse#(_x1)]] 
  [[shuffle#(cons(_x0, _x1))]] = 0 >= 0 = [[shuffle#(reverse(_x1))]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[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)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[shuffle(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[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 + 2F0(x1,x1) >= x1 = [[_x1]] 
  [[apply2(_F0, _x1, s(_x2))]] = x1 + 2F0(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 the dependency pair problem (P_2, R_1, static, formative) by (P_3, R_1, static, formative), where P_3 consists of:

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

Thus, the original system is terminating if (P_3, R_1, static, formative) is finite.

We consider the dependency pair problem (P_3, R_1, static, 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:

  reverse#(cons(X, Y)) >? reverse#(Y) 
  shuffle#(cons(X, Y)) >? shuffle#(reverse(Y)) 
  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 + 2G0(y1,y2) 
  cons = \y0y1.2 + y1 
  fst = \y0.y0 
  nil = 0 
  p = \y0y1.y0 
  prefixshuffle# = \y0y1.0 
  pshuffle = \y0.0 
  reverse = \y0.y0 
  reverse# = \y0.y0 
  s = \y0.0 
  shuffle = \y0.y0 
  shuffle# = \y0.0 

Using this interpretation, the requirements translate to:

  [[reverse#(cons(_x0, _x1))]] = 2 + x1 > x1 = [[reverse#(_x1)]] 
  [[shuffle#(cons(_x0, _x1))]] = 0 >= 0 = [[shuffle#(reverse(_x1))]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[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)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[shuffle(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[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 + 2F0(x1,0) >= x1 = [[_x1]] 
  [[apply2(_F0, _x1, s(_x2))]] = x1 + 2F0(x1,0) >= F0(x1,0) = [[_F0 _x1 s(_x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_1, static, formative) by (P_4, R_1, static, formative), where P_4 consists of:

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

Thus, the original system is terminating if (P_4, R_1, static, formative) is finite.

We consider the dependency pair problem (P_4, R_1, static, 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:

  shuffle#(cons(X, Y)) >? shuffle#(reverse(Y)) 
  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,y2) 
  cons = \y0y1.1 + y1 
  fst = \y0.y0 
  nil = 0 
  p = \y0y1.y0 
  prefixshuffle# = \y0y1.0 
  pshuffle = \y0.0 
  reverse = \y0.y0 
  s = \y0.0 
  shuffle = \y0.y0 
  shuffle# = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[shuffle#(cons(_x0, _x1))]] = 2 + 2x1 > 2x1 = [[shuffle#(reverse(_x1))]] 
  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 0 >= 0 = [[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))]] = 1 + x1 >= 1 + x1 = [[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(x1,0) >= x1 = [[_x1]] 
  [[apply2(_F0, _x1, s(_x2))]] = x1 + F0(x1,0) >= F0(x1,0) = [[_F0 _x1 s(_x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_1, static, formative) by (P_5, R_1, static, formative), where P_5 consists of:

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

Thus, the original system is terminating if (P_5, R_1, static, formative) is finite.

We consider the dependency pair problem (P_5, R_1, static, 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 = \y0y1.2 + y1 
  fst = \y0.y0 
  nil = 0 
  p = \y0y1.y0 
  prefixshuffle# = \y0y1.2y1 
  pshuffle = \y0.0 
  reverse = \y0.y0 
  s = \y0.0 
  shuffle = \y0.y0 

Using this interpretation, the requirements translate to:

  [[prefixshuffle#(_x0, cons(_x1, _x2))]] = 4 + 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)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[shuffle(nil)]] = 0 >= 0 = [[nil]] 
  [[shuffle(cons(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[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_5, 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.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[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.