We consider the system qsort.

  Alphabet:

    0 : [] --> nat 
    app : [list * list] --> list 
    cons : [nat * list] --> list 
    false : [] --> bool 
    filter : [nat -> bool * list] --> list 
    gr : [nat * nat] --> bool 
    if : [bool * list * list] --> list 
    le : [nat * nat] --> bool 
    nil : [] --> list 
    qsort : [list] --> list 
    s : [nat] --> nat 
    true : [] --> bool 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    app(nil, x) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    gr(0, x) => false 
    gr(s(x), 0) => true 
    gr(s(x), s(y)) => gr(x, y) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, y)), filter(f, y)) 
    qsort(nil) => nil 
    qsort(cons(x, y)) => app(qsort(filter(/\z.le(z, x), y)), app(cons(x, nil), qsort(filter(/\u.gr(u, x), 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:

  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  app(nil, X) => X 
  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  gr(0, X) => false 
  gr(s(X), 0) => true 
  gr(s(X), s(Y)) => gr(X, 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) RisEmptyProof [EQUIVALENT]
 || (4) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    gr(0, %X) -> false
 ||    gr(s(%X), 0) -> true
 ||    gr(s(%X), s(%Y)) -> gr(%X, %Y)
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Knuth-Bendix order [KBO] with precedence:s_1 > gr_2 > if_3 > 0 > le_2 > app_2 > cons_2 > nil > false > true
 || 
 || and weight map:
 || 
 ||    true=2
 ||    false=2
 ||    nil=1
 ||    0=1
 ||    s_1=0
 ||    if_3=0
 ||    app_2=0
 ||    cons_2=0
 ||    le_2=0
 ||    gr_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:
 || 
 ||    if(true, %X, %Y) -> %X
 ||    if(false, %X, %Y) -> %Y
 ||    app(nil, %X) -> %X
 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    gr(0, %X) -> false
 ||    gr(s(%X), 0) -> true
 ||    gr(s(%X), s(%Y)) -> gr(%X, %Y)
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (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] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
    1] filter#(F, cons(X, Y)) =#> F(X)   
    2] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    3] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    4] qsort#(cons(X, Y)) =#> app#(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y))))   
    5] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
    6] qsort#(cons(X, Y)) =#> filter#(/\x.le(x, X), Y)   
    7] qsort#(cons(X, Y)) =#> le#(x, X)   
    8] qsort#(cons(X, Y)) =#> app#(cons(X, nil), qsort(filter(/\x.gr(x, X), Y)))   
    9] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, X), Y))   
    10] qsort#(cons(X, Y)) =#> filter#(/\x.gr(x, X), Y)   
    11] qsort#(cons(X, Y)) =#> gr#(x, X)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    app(nil, X) => X 
    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    gr(0, X) => false 
    gr(s(X), 0) => true 
    gr(s(X), s(Y)) => gr(X, Y) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
    qsort(nil) => nil 
    qsort(cons(X, Y)) => app(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), 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 :  
    * 1 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 
    * 2 : 0, 1, 2, 3 
    * 3 : 0, 1, 2, 3 
    * 4 :  
    * 5 : 4, 5, 6, 7, 8, 9, 10, 11 
    * 6 : 0, 1, 2, 3 
    * 7 :  
    * 8 :  
    * 9 : 4, 5, 6, 7, 8, 9, 10, 11 
    * 10 : 0, 1, 2, 3 
    * 11 :  

This graph has the following strongly connected components:

  P_1:

    filter#(F, cons(X, Y)) =#> F(X)   
    filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
    qsort#(cons(X, Y)) =#> filter#(/\x.le(x, X), Y)   
    qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, X), Y))   
    qsort#(cons(X, Y)) =#> filter#(/\x.gr(x, X), 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:

  filter#(F, cons(X, Y)) >? F(X) 
  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  qsort#(cons(X, Y)) >? qsort#(filter(/\x.le-(x, X), Y)) 
  qsort#(cons(X, Y)) >? filter#(/\x.le-(x, X), Y) 
  qsort#(cons(X, Y)) >? qsort#(filter(/\x.gr-(x, X), Y)) 
  qsort#(cons(X, Y)) >? filter#(/\x.gr-(x, X), Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  gr(0, X) >= false 
  gr(s(X), 0) >= true 
  gr(s(X), s(Y)) >= gr(X, Y) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
  qsort(nil) >= nil 
  qsort(cons(X, Y)) >= app(qsort(filter(/\x.le-(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr-(y, X), Y)))) 
  gr-(X, Y) >= gr(X, Y) 
  le-(X, Y) >= le(X, Y) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[false]] = _|_ 
  [[filter(x_1, x_2)]] = filter(x_2) 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[le(x_1, x_2)]] = le(x_1) 
  [[nil]] = _|_ 
  [[true]] = _|_ 

We choose Lex = {app} and Mul = {0, @_{o -> o}, cons, filter, filter#, gr, gr-, if, le, le-, qsort, qsort#, s}, and the following precedence: 0 > qsort > app > cons = filter > qsort# > gr- > gr > if > le- > le = s > filter# > @_{o -> o}

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  filter#(F, cons(X, Y)) > @_{o -> o}(F, X) 
  filter#(F, cons(X, Y)) >= filter#(F, Y) 
  filter#(F, cons(X, Y)) >= filter#(F, Y) 
  qsort#(cons(X, Y)) >= qsort#(filter(Y)) 
  qsort#(cons(X, Y)) >= filter#(/\x.le-(x, X), Y) 
  qsort#(cons(X, Y)) > qsort#(filter(Y)) 
  qsort#(cons(X, Y)) >= filter#(/\x.gr-(x, X), Y) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  app(_|_, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  le(0) >= _|_ 
  le(s(X)) >= _|_ 
  le(s(X)) >= le(X) 
  gr(0, X) >= _|_ 
  gr(s(X), 0) >= _|_ 
  gr(s(X), s(Y)) >= gr(X, Y) 
  filter(_|_) >= _|_ 
  filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y)) 
  qsort(_|_) >= _|_ 
  qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y)))) 
  gr-(X, Y) >= gr(X, Y) 
  le-(X, Y) >= le(X) 

With these choices, we have:

  1] filter#(F, cons(X, Y)) > @_{o -> o}(F, X)  because [2], by definition 
  2] filter#*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter# > @_{o -> o}, [3] and [5], by (Copy) 
  3] filter#*(F, cons(X, Y)) >= F  because [4], by (Select) 
  4] F >= F  by (Meta) 
  5] filter#*(F, cons(X, Y)) >= X  because [6], by (Select) 
  6] cons(X, Y) >= X  because [7], by (Star) 
  7] cons*(X, Y) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 

  9] filter#(F, cons(X, Y)) >= filter#(F, Y)  because [10], by (Star) 
  10] filter#*(F, cons(X, Y)) >= filter#(F, Y)  because filter# in Mul, [11] and [12], by (Stat) 
  11] F >= F  by (Meta) 
  12] cons(X, Y) > Y  because [13], by definition 
  13] cons*(X, Y) >= Y  because [14], by (Select) 
  14] Y >= Y  by (Meta) 

  15] filter#(F, cons(X, Y)) >= filter#(F, Y)  because [10], by (Star) 

  16] qsort#(cons(X, Y)) >= qsort#(filter(Y))  because [17], by (Star) 
  17] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because qsort# in Mul and [18], by (Stat) 
  18] cons(X, Y) > filter(Y)  because [19], by definition 
  19] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [20], by (Stat) 
  20] Y >= Y  by (Meta) 

  21] qsort#(cons(X, Y)) >= filter#(/\x.le-(x, X), Y)  because [22], by (Star) 
  22] qsort#*(cons(X, Y)) >= filter#(/\x.le-(x, X), Y)  because qsort# > filter#, [23] and [31], by (Copy) 
  23] qsort#*(cons(X, Y)) >= /\y.le-(y, X)  because [24], by (F-Abs) 
  24] qsort#*(cons(X, Y), x) >= le-(x, X)  because qsort# > le-, [25] and [27], by (Copy) 
  25] qsort#*(cons(X, Y), x) >= x  because [26], by (Select) 
  26] x >= x  by (Var) 
  27] qsort#*(cons(X, Y), x) >= X  because [28], by (Select) 
  28] cons(X, Y) >= X  because [29], by (Star) 
  29] cons*(X, Y) >= X  because [30], by (Select) 
  30] X >= X  by (Meta) 
  31] qsort#*(cons(X, Y)) >= Y  because [32], by (Select) 
  32] cons(X, Y) >= Y  because [33], by (Star) 
  33] cons*(X, Y) >= Y  because [20], by (Select) 

  34] qsort#(cons(X, Y)) > qsort#(filter(Y))  because [35], by definition 
  35] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because [36], by (Select) 
  36] cons(X, Y) >= qsort#(filter(Y))  because [37], by (Star) 
  37] cons*(X, Y) >= qsort#(filter(Y))  because cons > qsort# and [38], by (Copy) 
  38] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [20], by (Stat) 

  39] qsort#(cons(X, Y)) >= filter#(/\x.gr-(x, X), Y)  because [40], by (Star) 
  40] qsort#*(cons(X, Y)) >= filter#(/\x.gr-(x, X), Y)  because qsort# > filter#, [41] and [31], by (Copy) 
  41] qsort#*(cons(X, Y)) >= /\y.gr-(y, X)  because [42], by (F-Abs) 
  42] qsort#*(cons(X, Y), x) >= gr-(x, X)  because qsort# > gr-, [43] and [45], by (Copy) 
  43] qsort#*(cons(X, Y), x) >= x  because [44], by (Select) 
  44] x >= x  by (Var) 
  45] qsort#*(cons(X, Y), x) >= X  because [28], by (Select) 

  46] if(X, Y) >= X  because [47], by (Star) 
  47] if*(X, Y) >= X  because [48], by (Select) 
  48] X >= X  by (Meta) 

  49] if(X, Y) >= Y  because [50], by (Star) 
  50] if*(X, Y) >= Y  because [51], by (Select) 
  51] Y >= Y  by (Meta) 

  52] app(_|_, X) >= X  because [53], by (Star) 
  53] app*(_|_, X) >= X  because [54], by (Select) 
  54] X >= X  by (Meta) 

  55] app(cons(X, Y), Z) >= cons(X, app(Y, Z))  because [56], by (Star) 
  56] app*(cons(X, Y), Z) >= cons(X, app(Y, Z))  because app > cons, [57] and [61], by (Copy) 
  57] app*(cons(X, Y), Z) >= X  because [58], by (Select) 
  58] cons(X, Y) >= X  because [59], by (Star) 
  59] cons*(X, Y) >= X  because [60], by (Select) 
  60] X >= X  by (Meta) 
  61] app*(cons(X, Y), Z) >= app(Y, Z)  because [62], [65] and [67], by (Stat) 
  62] cons(X, Y) > Y  because [63], by definition 
  63] cons*(X, Y) >= Y  because [64], by (Select) 
  64] Y >= Y  by (Meta) 
  65] app*(cons(X, Y), Z) >= Y  because [66], by (Select) 
  66] cons(X, Y) >= Y  because [63], by (Star) 
  67] app*(cons(X, Y), Z) >= Z  because [68], by (Select) 
  68] Z >= Z  by (Meta) 

  69] le(0) >= _|_  by (Bot) 

  70] le(s(X)) >= _|_  by (Bot) 

  71] le(s(X)) >= le(X)  because [72], by (Star) 
  72] le*(s(X)) >= le(X)  because [73], by (Select) 
  73] s(X) >= le(X)  because s = le, s in Mul and [74], by (Fun) 
  74] X >= X  by (Meta) 

  75] gr(0, X) >= _|_  by (Bot) 

  76] gr(s(X), 0) >= _|_  by (Bot) 

  77] gr(s(X), s(Y)) >= gr(X, Y)  because [78], by (Star) 
  78] gr*(s(X), s(Y)) >= gr(X, Y)  because gr in Mul, [79] and [82], by (Stat) 
  79] s(X) >= X  because [80], by (Star) 
  80] s*(X) >= X  because [81], by (Select) 
  81] X >= X  by (Meta) 
  82] s(Y) > Y  because [83], by definition 
  83] s*(Y) >= Y  because [84], by (Select) 
  84] Y >= Y  by (Meta) 

  85] filter(_|_) >= _|_  by (Bot) 

  86] filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because [87], by (Star) 
  87] filter*(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because filter > if, [88] and [94], by (Copy) 
  88] filter*(cons(X, Y)) >= cons(X, filter(Y))  because filter = cons, filter in Mul, [89] and [91], by (Stat) 
  89] cons(X, Y) > X  because [90], by definition 
  90] cons*(X, Y) >= X  because [8], by (Select) 
  91] cons(X, Y) > filter(Y)  because [92], by definition 
  92] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [93], by (Stat) 
  93] Y >= Y  by (Meta) 
  94] filter*(cons(X, Y)) >= filter(Y)  because filter in Mul and [12], by (Stat) 

  95] qsort(_|_) >= _|_  by (Bot) 

  96] qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because [97], by (Star) 
  97] qsort*(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because qsort > app, [98] and [99], by (Copy) 
  98] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [18], by (Stat) 
  99] qsort*(cons(X, Y)) >= app(cons(X, _|_), qsort(filter(Y)))  because qsort > app, [100] and [104], by (Copy) 
  100] qsort*(cons(X, Y)) >= cons(X, _|_)  because [101], by (Select) 
  101] cons(X, Y) >= cons(X, _|_)  because cons in Mul, [102] and [103], by (Fun) 
  102] X >= X  by (Meta) 
  103] Y >= _|_  by (Bot) 
  104] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [105], by (Stat) 
  105] cons(X, Y) > filter(Y)  because [38], by definition 

  106] gr-(X, Y) >= gr(X, Y)  because [107], by (Star) 
  107] gr-*(X, Y) >= gr(X, Y)  because gr- > gr, [108] and [110], by (Copy) 
  108] gr-*(X, Y) >= X  because [109], by (Select) 
  109] X >= X  by (Meta) 
  110] gr-*(X, Y) >= Y  because [111], by (Select) 
  111] Y >= Y  by (Meta) 

  112] le-(X, Y) >= le(X)  because [113], by (Star) 
  113] le-*(X, Y) >= le(X)  because le- > le and [114], by (Copy) 
  114] le-*(X, Y) >= X  because [115], by (Select) 
  115] X >= X  by (Meta) 

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:

  filter#(F, cons(X, Y)) =#> filter#(F, Y)   
  filter#(F, cons(X, Y)) =#> filter#(F, Y)   
  qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
  qsort#(cons(X, Y)) =#> filter#(/\x.le(x, X), Y)   
  qsort#(cons(X, Y)) =#> filter#(/\x.gr(x, X), 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 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 
    * 1 : 0, 1 
    * 2 : 2, 3, 4 
    * 3 : 0, 1 
    * 4 : 0, 1 

This graph has the following strongly connected components:

  P_3:

    filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    filter#(F, cons(X, Y)) =#> filter#(F, Y)   

  P_4:

    qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_2, R_0, m, f) by (P_3, R_0, m, f) and (P_4, R_0, m, f).

Thus, the original system is terminating if each of (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_4, R_0, 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:

  qsort#(cons(X, Y)) >? qsort#(filter(/\x.le(x, X), Y)) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  le(0, X) >= true 
  le(s(X), 0) >= false 
  le(s(X), s(Y)) >= le(X, Y) 
  gr(0, X) >= false 
  gr(s(X), 0) >= true 
  gr(s(X), s(Y)) >= gr(X, Y) 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
  qsort(nil) >= nil 
  qsort(cons(X, Y)) >= app(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y)))) 

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[false]] = _|_ 
  [[filter(x_1, x_2)]] = filter(x_2) 
  [[gr(x_1, x_2)]] = gr(x_1) 
  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
  [[nil]] = _|_ 
  [[true]] = _|_ 

We choose Lex = {} and Mul = {0, @_{o -> o}, app, cons, filter, gr, if, le, qsort, qsort#, s}, and the following precedence: 0 > @_{o -> o} > le > qsort > app > cons = filter > if > qsort# > gr > s

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  qsort#(cons(X, Y)) > qsort#(filter(Y)) 
  if(X, Y) >= X 
  if(X, Y) >= Y 
  app(_|_, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  le(0, X) >= _|_ 
  le(s(X), 0) >= _|_ 
  le(s(X), s(Y)) >= le(X, Y) 
  gr(0) >= _|_ 
  gr(s(X)) >= _|_ 
  gr(s(X)) >= gr(X) 
  filter(_|_) >= _|_ 
  filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y)) 
  qsort(_|_) >= _|_ 
  qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y)))) 

With these choices, we have:

  1] qsort#(cons(X, Y)) > qsort#(filter(Y))  because [2], by definition 
  2] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because [3], by (Select) 
  3] cons(X, Y) >= qsort#(filter(Y))  because [4], by (Star) 
  4] cons*(X, Y) >= qsort#(filter(Y))  because cons > qsort# and [5], by (Copy) 
  5] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [6], by (Stat) 
  6] Y >= Y  by (Meta) 

  7] if(X, Y) >= X  because [8], by (Star) 
  8] if*(X, Y) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 

  10] if(X, Y) >= Y  because [11], by (Star) 
  11] if*(X, Y) >= Y  because [12], by (Select) 
  12] Y >= Y  by (Meta) 

  13] app(_|_, X) >= X  because [14], by (Star) 
  14] app*(_|_, X) >= X  because [15], by (Select) 
  15] X >= X  by (Meta) 

  16] app(cons(X, Y), Z) >= cons(X, app(Y, Z))  because [17], by (Star) 
  17] app*(cons(X, Y), Z) >= cons(X, app(Y, Z))  because app > cons, [18] and [22], by (Copy) 
  18] app*(cons(X, Y), Z) >= X  because [19], by (Select) 
  19] cons(X, Y) >= X  because [20], by (Star) 
  20] cons*(X, Y) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 
  22] app*(cons(X, Y), Z) >= app(Y, Z)  because app in Mul, [23] and [26], by (Stat) 
  23] cons(X, Y) > Y  because [24], by definition 
  24] cons*(X, Y) >= Y  because [25], by (Select) 
  25] Y >= Y  by (Meta) 
  26] Z >= Z  by (Meta) 

  27] le(0, X) >= _|_  by (Bot) 

  28] le(s(X), 0) >= _|_  by (Bot) 

  29] le(s(X), s(Y)) >= le(X, Y)  because [30], by (Star) 
  30] le*(s(X), s(Y)) >= le(X, Y)  because le in Mul, [31] and [34], by (Stat) 
  31] s(X) >= X  because [32], by (Star) 
  32] s*(X) >= X  because [33], by (Select) 
  33] X >= X  by (Meta) 
  34] s(Y) > Y  because [35], by definition 
  35] s*(Y) >= Y  because [36], by (Select) 
  36] Y >= Y  by (Meta) 

  37] gr(0) >= _|_  by (Bot) 

  38] gr(s(X)) >= _|_  by (Bot) 

  39] gr(s(X)) >= gr(X)  because gr in Mul and [40], by (Fun) 
  40] s(X) >= X  because [41], by (Star) 
  41] s*(X) >= X  because [42], by (Select) 
  42] X >= X  by (Meta) 

  43] filter(_|_) >= _|_  by (Bot) 

  44] filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because [45], by (Star) 
  45] filter*(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because filter > if, [46] and [53], by (Copy) 
  46] filter*(cons(X, Y)) >= cons(X, filter(Y))  because filter = cons, filter in Mul, [47] and [50], by (Stat) 
  47] cons(X, Y) > X  because [48], by definition 
  48] cons*(X, Y) >= X  because [49], by (Select) 
  49] X >= X  by (Meta) 
  50] cons(X, Y) > filter(Y)  because [51], by definition 
  51] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [52], by (Stat) 
  52] Y >= Y  by (Meta) 
  53] filter*(cons(X, Y)) >= filter(Y)  because [54], by (Select) 
  54] cons(X, Y) >= filter(Y)  because [51], by (Star) 

  55] qsort(_|_) >= _|_  by (Bot) 

  56] qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because [57], by (Star) 
  57] qsort*(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because qsort > app, [58] and [60], by (Copy) 
  58] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [59], by (Stat) 
  59] cons(X, Y) > filter(Y)  because [5], by definition 
  60] qsort*(cons(X, Y)) >= app(cons(X, _|_), qsort(filter(Y)))  because qsort > app, [61] and [65], by (Copy) 
  61] qsort*(cons(X, Y)) >= cons(X, _|_)  because [62], by (Select) 
  62] cons(X, Y) >= cons(X, _|_)  because cons in Mul, [63] and [64], by (Fun) 
  63] X >= X  by (Meta) 
  64] Y >= _|_  by (Bot) 
  65] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [66], by (Stat) 
  66] cons(X, Y) > filter(Y)  because [67], by definition 
  67] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [6], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, R_0, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 
  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, 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.