2.85/2.09	YES
2.93/2.16	We consider the system theBenchmark.
2.93/2.16	
2.93/2.16	  Alphabet:
2.93/2.16	
2.93/2.16	    0 : [] --> nat 
2.93/2.16	    app : [list * list] --> list 
2.93/2.16	    cons : [nat * list] --> list 
2.93/2.16	    false : [] --> bool 
2.93/2.16	    filter : [nat -> bool * list] --> list 
2.93/2.16	    gr : [nat * nat] --> bool 
2.93/2.16	    if : [bool * list * list] --> list 
2.93/2.16	    le : [nat * nat] --> bool 
2.93/2.16	    nil : [] --> list 
2.93/2.16	    qsort : [list] --> list 
2.93/2.16	    s : [nat] --> nat 
2.93/2.16	    true : [] --> bool 
2.93/2.16	
2.93/2.16	  Rules:
2.93/2.16	
2.93/2.16	    if(true, x, y) => x 
2.93/2.16	    if(false, x, y) => y 
2.93/2.16	    app(nil, x) => x 
2.93/2.16	    app(cons(x, y), z) => cons(x, app(y, z)) 
2.93/2.16	    le(0, x) => true 
2.93/2.16	    le(s(x), 0) => false 
2.93/2.16	    le(s(x), s(y)) => le(x, y) 
2.93/2.16	    gr(0, x) => false 
2.93/2.16	    gr(s(x), 0) => true 
2.93/2.16	    gr(s(x), s(y)) => gr(x, y) 
2.93/2.16	    filter(f, nil) => nil 
2.93/2.16	    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, y)), filter(f, y)) 
2.93/2.16	    qsort(nil) => nil 
2.93/2.16	    qsort(cons(x, y)) => app(qsort(filter(/\z.le(z, x), y)), app(cons(x, nil), qsort(filter(/\u.gr(u, x), y)))) 
2.93/2.16	
2.93/2.16	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
2.93/2.16	
2.93/2.16	We observe that the rules contain a first-order subset:
2.93/2.16	
2.93/2.16	  if(true, X, Y) => X 
2.93/2.16	  if(false, X, Y) => Y 
2.93/2.16	  app(nil, X) => X 
2.93/2.16	  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
2.93/2.16	  le(0, X) => true 
2.93/2.16	  le(s(X), 0) => false 
2.93/2.16	  le(s(X), s(Y)) => le(X, Y) 
2.93/2.16	  gr(0, X) => false 
2.93/2.16	  gr(s(X), 0) => true 
2.93/2.16	  gr(s(X), s(Y)) => gr(X, Y) 
2.93/2.16	
2.93/2.16	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.
2.93/2.16	
2.93/2.16	According to the external first-order termination prover, this system is indeed terminating:
2.93/2.16	
2.93/2.16	 || proof of resources/system.trs
2.93/2.16	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
2.93/2.16	 || 
2.93/2.16	 || 
2.93/2.16	 || Termination w.r.t. Q of the given QTRS could be proven:
2.93/2.16	 || 
2.93/2.16	 || (0) QTRS
2.93/2.16	 || (1) QTRSRRRProof [EQUIVALENT]
2.93/2.16	 || (2) QTRS
2.93/2.16	 || (3) RisEmptyProof [EQUIVALENT]
2.93/2.16	 || (4) YES
2.93/2.16	 || 
2.93/2.16	 || 
2.93/2.16	 || ----------------------------------------
2.93/2.16	 || 
2.93/2.16	 || (0)
2.93/2.16	 || Obligation:
2.93/2.16	 || Q restricted rewrite system:
2.93/2.16	 || The TRS R consists of the following rules:
2.93/2.16	 || 
2.93/2.16	 ||    if(true, %X, %Y) -> %X
2.93/2.16	 ||    if(false, %X, %Y) -> %Y
2.93/2.16	 ||    app(nil, %X) -> %X
2.93/2.16	 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
2.93/2.16	 ||    le(0, %X) -> true
2.93/2.16	 ||    le(s(%X), 0) -> false
2.93/2.16	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
2.93/2.16	 ||    gr(0, %X) -> false
2.93/2.16	 ||    gr(s(%X), 0) -> true
2.93/2.16	 ||    gr(s(%X), s(%Y)) -> gr(%X, %Y)
2.93/2.16	 || 
2.93/2.16	 || Q is empty.
2.93/2.16	 || 
2.93/2.16	 || ----------------------------------------
2.93/2.16	 || 
2.93/2.16	 || (1) QTRSRRRProof (EQUIVALENT)
2.93/2.16	 || Used ordering:
2.93/2.16	 || Knuth-Bendix order [KBO] with precedence:s_1 > gr_2 > if_3 > 0 > le_2 > app_2 > cons_2 > nil > false > true
2.93/2.16	 || 
2.93/2.16	 || and weight map:
2.93/2.16	 || 
2.93/2.16	 ||    true=2
2.93/2.16	 ||    false=2
2.93/2.16	 ||    nil=1
2.93/2.16	 ||    0=1
2.93/2.16	 ||    s_1=0
2.93/2.16	 ||    if_3=0
2.93/2.16	 ||    app_2=0
2.93/2.16	 ||    cons_2=0
2.93/2.16	 ||    le_2=0
2.93/2.16	 ||    gr_2=0
2.93/2.16	 || 
2.93/2.16	 || The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
2.93/2.16	 || 
2.93/2.16	 ||    if(true, %X, %Y) -> %X
2.93/2.16	 ||    if(false, %X, %Y) -> %Y
2.93/2.16	 ||    app(nil, %X) -> %X
2.93/2.16	 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
2.93/2.16	 ||    le(0, %X) -> true
2.93/2.16	 ||    le(s(%X), 0) -> false
2.93/2.16	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
2.93/2.16	 ||    gr(0, %X) -> false
2.93/2.16	 ||    gr(s(%X), 0) -> true
2.93/2.16	 ||    gr(s(%X), s(%Y)) -> gr(%X, %Y)
2.93/2.16	 || 
2.93/2.16	 || 
2.93/2.16	 || 
2.93/2.16	 || 
2.93/2.16	 || ----------------------------------------
2.93/2.16	 || 
2.93/2.16	 || (2)
2.93/2.16	 || Obligation:
2.93/2.16	 || Q restricted rewrite system:
2.93/2.16	 || R is empty.
2.93/2.16	 || Q is empty.
2.93/2.16	 || 
2.93/2.16	 || ----------------------------------------
2.93/2.16	 || 
2.93/2.16	 || (3) RisEmptyProof (EQUIVALENT)
2.93/2.16	 || The TRS R is empty. Hence, termination is trivially proven.
2.93/2.16	 || ----------------------------------------
2.93/2.16	 || 
2.93/2.16	 || (4)
2.93/2.16	 || YES
2.93/2.16	 || 
2.93/2.16	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]).
2.93/2.16	
2.93/2.16	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
2.93/2.16	
2.93/2.16	  Dependency Pairs P_0:
2.93/2.16	
2.93/2.16	    0] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
2.93/2.16	    1] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
2.93/2.16	    2] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
2.93/2.16	    3] qsort#(cons(X, Y)) =#> app#(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y))))   
2.93/2.16	    4] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
2.93/2.16	    5] qsort#(cons(X, Y)) =#> filter#(/\x.le(x, X), Y)   
2.93/2.16	    6] qsort#(cons(X, Y)) =#> le#(Z, X)   
2.93/2.16	    7] qsort#(cons(X, Y)) =#> app#(cons(X, nil), qsort(filter(/\x.gr(x, X), Y)))   
2.93/2.16	    8] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, X), Y))   
2.93/2.16	    9] qsort#(cons(X, Y)) =#> filter#(/\x.gr(x, X), Y)   
2.93/2.16	    10] qsort#(cons(X, Y)) =#> gr#(Z, X)   
2.93/2.16	
2.93/2.16	  Rules R_0:
2.93/2.16	
2.93/2.16	    if(true, X, Y) => X 
2.93/2.16	    if(false, X, Y) => Y 
2.93/2.16	    app(nil, X) => X 
2.93/2.16	    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
2.93/2.16	    le(0, X) => true 
2.93/2.16	    le(s(X), 0) => false 
2.93/2.16	    le(s(X), s(Y)) => le(X, Y) 
2.93/2.16	    gr(0, X) => false 
2.93/2.16	    gr(s(X), 0) => true 
2.93/2.16	    gr(s(X), s(Y)) => gr(X, Y) 
2.93/2.16	    filter(F, nil) => nil 
2.93/2.16	    filter(F, cons(X, Y)) => if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
2.93/2.16	    qsort(nil) => nil 
2.93/2.16	    qsort(cons(X, Y)) => app(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y)))) 
2.93/2.16	
2.93/2.16	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
2.93/2.16	
2.93/2.16	We consider the dependency pair problem (P_0, R_0, static, formative).
2.93/2.16	
2.93/2.16	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
2.93/2.16	
2.93/2.16	    * 0 :  
2.93/2.16	    * 1 : 0, 1, 2 
2.93/2.16	    * 2 : 0, 1, 2 
2.93/2.16	    * 3 :  
2.93/2.16	    * 4 : 3, 4, 5, 6, 7, 8, 9, 10 
2.93/2.16	    * 5 : 0, 1, 2 
2.93/2.16	    * 6 :  
2.93/2.16	    * 7 :  
2.93/2.16	    * 8 : 3, 4, 5, 6, 7, 8, 9, 10 
2.93/2.16	    * 9 : 0, 1, 2 
2.93/2.16	    * 10 :  
2.93/2.16	
2.93/2.16	This graph has the following strongly connected components:
2.93/2.16	
2.93/2.16	  P_1:
2.93/2.16	
2.93/2.16	    filter#(F, cons(X, Y)) =#> filter#(F, Y)   
2.93/2.16	    filter#(F, cons(X, Y)) =#> filter#(F, Y)   
2.93/2.16	
2.93/2.16	  P_2:
2.93/2.16	
2.93/2.16	    qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
2.93/2.16	    qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, X), Y))   
2.93/2.16	
2.93/2.16	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).
2.93/2.16	
2.93/2.16	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.
2.93/2.16	
2.93/2.16	We consider the dependency pair problem (P_2, R_0, static, formative).
2.93/2.16	
2.93/2.16	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:
2.93/2.16	
2.93/2.16	  qsort#(cons(X, Y)) >? qsort#(filter(/\x.le(x, X), Y)) 
2.93/2.16	  qsort#(cons(X, Y)) >? qsort#(filter(/\x.gr(x, X), Y)) 
2.93/2.16	  if(true, X, Y) >= X 
2.93/2.16	  if(false, X, Y) >= Y 
2.93/2.16	  app(nil, X) >= X 
2.93/2.16	  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
2.93/2.16	  le(0, X) >= true 
2.93/2.16	  le(s(X), 0) >= false 
2.93/2.16	  le(s(X), s(Y)) >= le(X, Y) 
2.93/2.16	  gr(0, X) >= false 
2.93/2.16	  gr(s(X), 0) >= true 
2.93/2.16	  gr(s(X), s(Y)) >= gr(X, Y) 
2.93/2.16	  filter(F, nil) >= nil 
2.93/2.16	  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
2.93/2.16	  qsort(nil) >= nil 
2.93/2.16	  qsort(cons(X, Y)) >= app(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y)))) 
2.93/2.16	
2.93/2.16	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.)
2.93/2.16	
2.93/2.16	We use a recursive path ordering as defined in [Kop12, Chapter 5].
2.93/2.16	
2.93/2.16	Argument functions:
2.93/2.16	
2.93/2.16	  [[false]] = _|_ 
2.93/2.16	  [[filter(x_1, x_2)]] = filter(x_2) 
2.93/2.16	  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
2.93/2.16	  [[le(x_1, x_2)]] = x_2 
2.93/2.16	  [[nil]] = _|_ 
2.93/2.16	  [[true]] = _|_ 
2.93/2.16	
2.93/2.16	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
2.93/2.16	
2.93/2.16	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
2.93/2.16	
2.93/2.16	  qsort#(cons(X, Y)) >= qsort#(filter(Y)) 
2.93/2.16	  qsort#(cons(X, Y)) > qsort#(filter(Y)) 
2.93/2.16	  if(X, Y) >= X 
2.93/2.16	  if(X, Y) >= Y 
2.93/2.16	  app(_|_, X) >= X 
2.93/2.16	  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
2.93/2.16	  X >= _|_ 
2.93/2.16	  0 >= _|_ 
2.93/2.16	  s(X) >= X 
2.93/2.16	  gr(0, X) >= _|_ 
2.93/2.16	  gr(s(X), 0) >= _|_ 
2.93/2.16	  gr(s(X), s(Y)) >= gr(X, Y) 
2.93/2.16	  filter(_|_) >= _|_ 
2.93/2.16	  filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y)) 
2.93/2.16	  qsort(_|_) >= _|_ 
2.93/2.16	  qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y)))) 
2.93/2.16	
2.93/2.16	With these choices, we have:
2.93/2.16	
2.93/2.16	  1] qsort#(cons(X, Y)) >= qsort#(filter(Y))  because [2], by (Star) 
2.93/2.16	  2] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because qsort# in Mul and [3], by (Stat) 
2.93/2.16	  3] cons(X, Y) > filter(Y)  because [4], by definition 
2.93/2.16	  4] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [5], by (Stat) 
2.93/2.16	  5] Y >= Y  by (Meta) 
2.93/2.16	
2.93/2.16	  6] qsort#(cons(X, Y)) > qsort#(filter(Y))  because [7], by definition 
2.93/2.16	  7] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because qsort# in Mul and [8], by (Stat) 
2.93/2.16	  8] cons(X, Y) > filter(Y)  because [9], by definition 
2.93/2.16	  9] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [5], by (Stat) 
2.93/2.16	
2.93/2.16	  10] if(X, Y) >= X  because [11], by (Star) 
2.93/2.16	  11] if*(X, Y) >= X  because [12], by (Select) 
2.93/2.16	  12] X >= X  by (Meta) 
2.93/2.16	
2.93/2.16	  13] if(X, Y) >= Y  because [14], by (Star) 
2.93/2.16	  14] if*(X, Y) >= Y  because [15], by (Select) 
2.93/2.16	  15] Y >= Y  by (Meta) 
2.93/2.16	
2.93/2.16	  16] app(_|_, X) >= X  because [17], by (Star) 
2.93/2.16	  17] app*(_|_, X) >= X  because [18], by (Select) 
2.93/2.16	  18] X >= X  by (Meta) 
2.93/2.16	
2.93/2.16	  19] app(cons(X, Y), Z) >= cons(X, app(Y, Z))  because [20], by (Star) 
2.93/2.16	  20] app*(cons(X, Y), Z) >= cons(X, app(Y, Z))  because app > cons, [21] and [25], by (Copy) 
2.93/2.16	  21] app*(cons(X, Y), Z) >= X  because [22], by (Select) 
2.93/2.16	  22] cons(X, Y) >= X  because [23], by (Star) 
2.93/2.16	  23] cons*(X, Y) >= X  because [24], by (Select) 
2.93/2.16	  24] X >= X  by (Meta) 
2.93/2.16	  25] app*(cons(X, Y), Z) >= app(Y, Z)  because [26], [29] and [31], by (Stat) 
2.93/2.16	  26] cons(X, Y) > Y  because [27], by definition 
2.93/2.16	  27] cons*(X, Y) >= Y  because [28], by (Select) 
2.93/2.16	  28] Y >= Y  by (Meta) 
2.93/2.16	  29] app*(cons(X, Y), Z) >= Y  because [30], by (Select) 
2.93/2.16	  30] cons(X, Y) >= Y  because [27], by (Star) 
2.93/2.16	  31] app*(cons(X, Y), Z) >= Z  because [32], by (Select) 
2.93/2.16	  32] Z >= Z  by (Meta) 
2.93/2.16	
2.93/2.16	  33] X >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  34] 0 >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  35] s(X) >= X  because [36], by (Star) 
2.93/2.16	  36] s*(X) >= X  because [37], by (Select) 
2.93/2.16	  37] X >= X  by (Meta) 
2.93/2.16	
2.93/2.16	  38] gr(0, X) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  39] gr(s(X), 0) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  40] gr(s(X), s(Y)) >= gr(X, Y)  because [41], by (Star) 
2.93/2.16	  41] gr*(s(X), s(Y)) >= gr(X, Y)  because gr in Mul, [42] and [45], by (Stat) 
2.93/2.16	  42] s(X) >= X  because [43], by (Star) 
2.93/2.16	  43] s*(X) >= X  because [44], by (Select) 
2.93/2.16	  44] X >= X  by (Meta) 
2.93/2.16	  45] s(Y) > Y  because [46], by definition 
2.93/2.16	  46] s*(Y) >= Y  because [47], by (Select) 
2.93/2.16	  47] Y >= Y  by (Meta) 
2.93/2.16	
2.93/2.16	  48] filter(_|_) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  49] filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because [50], by (Star) 
2.93/2.16	  50] filter*(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because filter > if, [51] and [58], by (Copy) 
2.93/2.16	  51] filter*(cons(X, Y)) >= cons(X, filter(Y))  because filter = cons, filter in Mul, [52] and [55], by (Stat) 
2.93/2.16	  52] cons(X, Y) > X  because [53], by definition 
2.93/2.16	  53] cons*(X, Y) >= X  because [54], by (Select) 
2.93/2.16	  54] X >= X  by (Meta) 
2.93/2.16	  55] cons(X, Y) > filter(Y)  because [56], by definition 
2.93/2.16	  56] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [57], by (Stat) 
2.93/2.16	  57] Y >= Y  by (Meta) 
2.93/2.16	  58] filter*(cons(X, Y)) >= filter(Y)  because filter in Mul and [59], by (Stat) 
2.93/2.16	  59] cons(X, Y) > Y  because [60], by definition 
2.93/2.16	  60] cons*(X, Y) >= Y  because [57], by (Select) 
2.93/2.16	
2.93/2.16	  61] qsort(_|_) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  62] qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because [63], by (Star) 
2.93/2.16	  63] qsort*(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because qsort > app, [64] and [65], by (Copy) 
2.93/2.16	  64] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [3], by (Stat) 
2.93/2.16	  65] qsort*(cons(X, Y)) >= app(cons(X, _|_), qsort(filter(Y)))  because qsort > app, [66] and [72], by (Copy) 
2.93/2.16	  66] qsort*(cons(X, Y)) >= cons(X, _|_)  because qsort > cons, [67] and [71], by (Copy) 
2.93/2.16	  67] qsort*(cons(X, Y)) >= X  because [68], by (Select) 
2.93/2.16	  68] cons(X, Y) >= X  because [69], by (Star) 
2.93/2.16	  69] cons*(X, Y) >= X  because [70], by (Select) 
2.93/2.16	  70] X >= X  by (Meta) 
2.93/2.16	  71] qsort*(cons(X, Y)) >= _|_  by (Bot) 
2.93/2.16	  72] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [8], by (Stat) 
2.93/2.16	
2.93/2.16	By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of:
2.93/2.16	
2.93/2.16	  qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y))   
2.93/2.16	
2.93/2.16	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite.
2.93/2.16	
2.93/2.16	We consider the dependency pair problem (P_3, R_0, static, formative).
2.93/2.16	
2.93/2.16	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:
2.93/2.16	
2.93/2.16	  qsort#(cons(X, Y)) >? qsort#(filter(/\x.le(x, X), Y)) 
2.93/2.16	  if(true, X, Y) >= X 
2.93/2.16	  if(false, X, Y) >= Y 
2.93/2.16	  app(nil, X) >= X 
2.93/2.16	  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
2.93/2.16	  le(0, X) >= true 
2.93/2.16	  le(s(X), 0) >= false 
2.93/2.16	  le(s(X), s(Y)) >= le(X, Y) 
2.93/2.16	  gr(0, X) >= false 
2.93/2.16	  gr(s(X), 0) >= true 
2.93/2.16	  gr(s(X), s(Y)) >= gr(X, Y) 
2.93/2.16	  filter(F, nil) >= nil 
2.93/2.16	  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
2.93/2.16	  qsort(nil) >= nil 
2.93/2.16	  qsort(cons(X, Y)) >= app(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y)))) 
2.93/2.16	
2.93/2.16	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.)
2.93/2.16	
2.93/2.16	We use a recursive path ordering as defined in [Kop12, Chapter 5].
2.93/2.16	
2.93/2.16	Argument functions:
2.93/2.16	
2.93/2.16	  [[false]] = _|_ 
2.93/2.16	  [[filter(x_1, x_2)]] = filter(x_2) 
2.93/2.16	  [[gr(x_1, x_2)]] = gr(x_1) 
2.93/2.16	  [[if(x_1, x_2, x_3)]] = if(x_2, x_3) 
2.93/2.16	  [[nil]] = _|_ 
2.93/2.16	  [[true]] = _|_ 
2.93/2.16	
2.93/2.16	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
2.93/2.16	
2.93/2.16	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
2.93/2.16	
2.93/2.16	  qsort#(cons(X, Y)) > qsort#(filter(Y)) 
2.93/2.16	  if(X, Y) >= X 
2.93/2.16	  if(X, Y) >= Y 
2.93/2.16	  app(_|_, X) >= X 
2.93/2.16	  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
2.93/2.16	  le(0, X) >= _|_ 
2.93/2.16	  le(s(X), 0) >= _|_ 
2.93/2.16	  le(s(X), s(Y)) >= le(X, Y) 
2.93/2.16	  gr(0) >= _|_ 
2.93/2.16	  gr(s(X)) >= _|_ 
2.93/2.16	  gr(s(X)) >= gr(X) 
2.93/2.16	  filter(_|_) >= _|_ 
2.93/2.16	  filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y)) 
2.93/2.16	  qsort(_|_) >= _|_ 
2.93/2.16	  qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y)))) 
2.93/2.16	
2.93/2.16	With these choices, we have:
2.93/2.16	
2.93/2.16	  1] qsort#(cons(X, Y)) > qsort#(filter(Y))  because [2], by definition 
2.93/2.16	  2] qsort#*(cons(X, Y)) >= qsort#(filter(Y))  because [3], by (Select) 
2.93/2.16	  3] cons(X, Y) >= qsort#(filter(Y))  because [4], by (Star) 
2.93/2.16	  4] cons*(X, Y) >= qsort#(filter(Y))  because cons > qsort# and [5], by (Copy) 
2.93/2.16	  5] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [6], by (Stat) 
2.93/2.16	  6] Y >= Y  by (Meta) 
2.93/2.16	
2.93/2.16	  7] if(X, Y) >= X  because [8], by (Star) 
2.93/2.16	  8] if*(X, Y) >= X  because [9], by (Select) 
2.93/2.16	  9] X >= X  by (Meta) 
2.93/2.16	
2.93/2.16	  10] if(X, Y) >= Y  because [11], by (Star) 
2.93/2.16	  11] if*(X, Y) >= Y  because [12], by (Select) 
2.93/2.16	  12] Y >= Y  by (Meta) 
2.93/2.16	
2.93/2.16	  13] app(_|_, X) >= X  because [14], by (Star) 
2.93/2.16	  14] app*(_|_, X) >= X  because [15], by (Select) 
2.93/2.16	  15] X >= X  by (Meta) 
2.93/2.16	
2.93/2.16	  16] app(cons(X, Y), Z) >= cons(X, app(Y, Z))  because [17], by (Star) 
2.93/2.16	  17] app*(cons(X, Y), Z) >= cons(X, app(Y, Z))  because app > cons, [18] and [22], by (Copy) 
2.93/2.16	  18] app*(cons(X, Y), Z) >= X  because [19], by (Select) 
2.93/2.16	  19] cons(X, Y) >= X  because [20], by (Star) 
2.93/2.16	  20] cons*(X, Y) >= X  because [21], by (Select) 
2.93/2.16	  21] X >= X  by (Meta) 
2.93/2.16	  22] app*(cons(X, Y), Z) >= app(Y, Z)  because app in Mul, [23] and [26], by (Stat) 
2.93/2.16	  23] cons(X, Y) > Y  because [24], by definition 
2.93/2.16	  24] cons*(X, Y) >= Y  because [25], by (Select) 
2.93/2.16	  25] Y >= Y  by (Meta) 
2.93/2.16	  26] Z >= Z  by (Meta) 
2.93/2.16	
2.93/2.16	  27] le(0, X) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  28] le(s(X), 0) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  29] le(s(X), s(Y)) >= le(X, Y)  because [30], by (Star) 
2.93/2.16	  30] le*(s(X), s(Y)) >= le(X, Y)  because le in Mul, [31] and [34], by (Stat) 
2.93/2.16	  31] s(X) >= X  because [32], by (Star) 
2.93/2.16	  32] s*(X) >= X  because [33], by (Select) 
2.93/2.16	  33] X >= X  by (Meta) 
2.93/2.16	  34] s(Y) > Y  because [35], by definition 
2.93/2.16	  35] s*(Y) >= Y  because [36], by (Select) 
2.93/2.16	  36] Y >= Y  by (Meta) 
2.93/2.16	
2.93/2.16	  37] gr(0) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  38] gr(s(X)) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  39] gr(s(X)) >= gr(X)  because gr in Mul and [40], by (Fun) 
2.93/2.16	  40] s(X) >= X  because [41], by (Star) 
2.93/2.16	  41] s*(X) >= X  because [42], by (Select) 
2.93/2.16	  42] X >= X  by (Meta) 
2.93/2.16	
2.93/2.16	  43] filter(_|_) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  44] filter(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because [45], by (Star) 
2.93/2.16	  45] filter*(cons(X, Y)) >= if(cons(X, filter(Y)), filter(Y))  because filter > if, [46] and [53], by (Copy) 
2.93/2.16	  46] filter*(cons(X, Y)) >= cons(X, filter(Y))  because filter = cons, filter in Mul, [47] and [50], by (Stat) 
2.93/2.16	  47] cons(X, Y) > X  because [48], by definition 
2.93/2.16	  48] cons*(X, Y) >= X  because [49], by (Select) 
2.93/2.16	  49] X >= X  by (Meta) 
2.93/2.16	  50] cons(X, Y) > filter(Y)  because [51], by definition 
2.93/2.16	  51] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [52], by (Stat) 
2.93/2.16	  52] Y >= Y  by (Meta) 
2.93/2.16	  53] filter*(cons(X, Y)) >= filter(Y)  because [54], by (Select) 
2.93/2.16	  54] cons(X, Y) >= filter(Y)  because [51], by (Star) 
2.93/2.16	
2.93/2.16	  55] qsort(_|_) >= _|_  by (Bot) 
2.93/2.16	
2.93/2.16	  56] qsort(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because [57], by (Star) 
2.93/2.16	  57] qsort*(cons(X, Y)) >= app(qsort(filter(Y)), app(cons(X, _|_), qsort(filter(Y))))  because qsort > app, [58] and [60], by (Copy) 
2.93/2.16	  58] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [59], by (Stat) 
2.93/2.16	  59] cons(X, Y) > filter(Y)  because [5], by definition 
2.93/2.16	  60] qsort*(cons(X, Y)) >= app(cons(X, _|_), qsort(filter(Y)))  because qsort > app, [61] and [65], by (Copy) 
2.93/2.16	  61] qsort*(cons(X, Y)) >= cons(X, _|_)  because [62], by (Select) 
2.93/2.16	  62] cons(X, Y) >= cons(X, _|_)  because cons in Mul, [63] and [64], by (Fun) 
2.93/2.16	  63] X >= X  by (Meta) 
2.93/2.16	  64] Y >= _|_  by (Bot) 
2.93/2.16	  65] qsort*(cons(X, Y)) >= qsort(filter(Y))  because qsort in Mul and [66], by (Stat) 
2.93/2.16	  66] cons(X, Y) > filter(Y)  because [67], by definition 
2.93/2.16	  67] cons*(X, Y) >= filter(Y)  because cons = filter, cons in Mul and [6], by (Stat) 
2.93/2.16	
2.93/2.16	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.
2.93/2.16	
2.93/2.16	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
2.93/2.16	
2.93/2.16	We consider the dependency pair problem (P_1, R_0, static, formative).
2.93/2.16	
2.93/2.16	We apply the subterm criterion with the following projection function:
2.93/2.16	
2.93/2.16	  nu(filter#) = 2 
2.93/2.16	
2.93/2.16	Thus, we can orient the dependency pairs as follows:
2.93/2.16	
2.93/2.16	  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 
2.93/2.16	  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 
2.93/2.16	
2.93/2.16	By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
2.93/2.16	
2.93/2.16	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
2.93/2.16	
2.93/2.16	
2.93/2.16	+++ Citations +++
2.93/2.16	
2.93/2.16	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
2.93/2.16	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
2.93/2.16	[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.
2.93/2.16	EOF