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


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YES
We consider the system theBenchmark.

  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) QTRSRRRProof [EQUIVALENT]
 || (4) QTRS
 || (5) RisEmptyProof [EQUIVALENT]
 || (6) 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:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    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(false) = 0
 ||    POL(gr(x_1, x_2)) = x_1 + x_2
 ||    POL(if(x_1, x_2, x_3)) = 2 + x_1 + x_2 + x_3
 ||    POL(le(x_1, x_2)) = 2 + x_1 + x_2
 ||    POL(nil) = 2
 ||    POL(s(x_1)) = 1 + 2*x_1
 ||    POL(true) = 1
 || With 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(s(%X), s(%Y)) -> gr(%X, %Y)
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    gr(0, %X) -> false
 ||    gr(s(%X), 0) -> true
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (3) QTRSRRRProof (EQUIVALENT)
 || Used ordering:
 || Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 1
 ||    POL(false) = 2
 ||    POL(gr(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
 ||    POL(s(x_1)) = 2 + x_1
 ||    POL(true) = 1
 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
 || 
 ||    gr(0, %X) -> false
 ||    gr(s(%X), 0) -> true
 || 
 || 
 || 
 || 
 || ----------------------------------------
 || 
 || (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 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] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
    1] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    2] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    3] qsort#(cons(X, Y)) =#> app#(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y))))   
    4] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
    5] qsort#(cons(X, Y)) =#> filter#(/\x.le(x, X), Y)   
    6] qsort#(cons(X, Y)) =#> le#(Z, X)   
    7] qsort#(cons(X, Y)) =#> app#(cons(X, nil), qsort(filter(/\x.gr(x, X), Y)))   
    8] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, X), Y))   
    9] qsort#(cons(X, Y)) =#> filter#(/\x.gr(x, X), Y)   
    10] qsort#(cons(X, Y)) =#> gr#(Z, 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 
    * 2 : 0, 1, 2 
    * 3 :  
    * 4 : 3, 4, 5, 6, 7, 8, 9, 10 
    * 5 : 0, 1, 2 
    * 6 :  
    * 7 :  
    * 8 : 3, 4, 5, 6, 7, 8, 9, 10 
    * 9 : 0, 1, 2 
    * 10 :  

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
    qsort#(cons(X, Y)) =#> qsort#(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) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, 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)) 
  qsort#(cons(X, Y)) >? qsort#(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)))) 

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)]] = x_2 
  [[nil]] = _|_ 
  [[true]] = _|_ 

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

With these choices, we have:

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

  6] qsort#(cons(X, Y)) > qsort#(filter(Y))  because [7], by definition 
  7] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because qsort# in Mul and [8], by (Stat) 
  8] cons(X, Y) > filter(Y)  because [9], by definition 
  9] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [5], by (Stat) 

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

  13] if(X, Y) >= Y  because [14], by (Star) 
  14] if*(X, Y) >= Y  because [15], by (Select) 
  15] Y >= Y  by (Meta) 

  16] app(_|_, X) >= X  because [17], by (Star) 
  17] app*(_|_, X) >= X  because [18], by (Select) 
  18] X >= X  by (Meta) 

  19] app(cons(X, Y), Z) >= cons(X, app(Y, Z))  because [20], by (Star) 
  20] app*(cons(X, Y), Z) >= cons(X, app(Y, Z))  because app > cons, [21] and [25], by (Copy) 
  21] app*(cons(X, Y), Z) >= X  because [22], by (Select) 
  22] cons(X, Y) >= X  because [23], by (Star) 
  23] cons*(X, Y) >= X  because [24], by (Select) 
  24] X >= X  by (Meta) 
  25] app*(cons(X, Y), Z) >= app(Y, Z)  because [26], [29] and [31], by (Stat) 
  26] cons(X, Y) > Y  because [27], by definition 
  27] cons*(X, Y) >= Y  because [28], by (Select) 
  28] Y >= Y  by (Meta) 
  29] app*(cons(X, Y), Z) >= Y  because [30], by (Select) 
  30] cons(X, Y) >= Y  because [27], by (Star) 
  31] app*(cons(X, Y), Z) >= Z  because [32], by (Select) 
  32] Z >= Z  by (Meta) 

  33] X >= _|_  by (Bot) 

  34] 0 >= _|_  by (Bot) 

  35] s(X) >= X  because [36], by (Star) 
  36] s*(X) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 

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

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

  40] gr(s(X), s(Y)) >= gr(X, Y)  because [41], by (Star) 
  41] gr*(s(X), s(Y)) >= gr(X, Y)  because gr in Mul, [42] and [45], by (Stat) 
  42] s(X) >= X  because [43], by (Star) 
  43] s*(X) >= X  because [44], by (Select) 
  44] X >= X  by (Meta) 
  45] s(Y) > Y  because [46], by definition 
  46] s*(Y) >= Y  because [47], by (Select) 
  47] Y >= Y  by (Meta) 

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

  49] filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because [50], by (Star) 
  50] filter*(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because filter > if, [51] and [58], by (Copy) 
  51] filter*(cons(X, Y)) >= cons(X, filter(Y))  because filter = cons, filter in Mul, [52] and [55], by (Stat) 
  52] cons(X, Y) > X  because [53], by definition 
  53] cons*(X, Y) >= X  because [54], by (Select) 
  54] X >= X  by (Meta) 
  55] cons(X, Y) > filter(Y)  because [56], by definition 
  56] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [57], by (Stat) 
  57] Y >= Y  by (Meta) 
  58] filter*(cons(X, Y)) >= filter(Y)  because filter in Mul and [59], by (Stat) 
  59] cons(X, Y) > Y  because [60], by definition 
  60] cons*(X, Y) >= Y  because [57], by (Select) 

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

  62] qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because [63], by (Star) 
  63] qsort*(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because qsort > app, [64] and [65], by (Copy) 
  64] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [3], by (Stat) 
  65] qsort*(cons(X, Y)) >= app(cons(X, _|_), qsort(filter(Y)))  because qsort > app, [66] and [72], by (Copy) 
  66] qsort*(cons(X, Y)) >= cons(X, _|_)  because qsort > cons, [67] and [71], by (Copy) 
  67] qsort*(cons(X, Y)) >= X  because [68], by (Select) 
  68] cons(X, Y) >= X  because [69], by (Star) 
  69] cons*(X, Y) >= X  because [70], by (Select) 
  70] X >= X  by (Meta) 
  71] qsort*(cons(X, Y)) >= _|_  by (Bot) 
  72] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [8], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_3, R_0, minimal, formative), where P_3 consists of:

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

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_3, 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_3, 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_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, 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 [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, 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 +++

[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.