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


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


YES
We consider the system theBenchmark.

  Alphabet:

    0 : [] --> b 
    add : [b * c] --> c 
    app : [c * c] --> c 
    false : [] --> a 
    filter : [b -> a * c] --> c 
    filter2 : [a * b -> a * b * c] --> c 
    high : [b * c] --> c 
    if!fac6220high : [a * b * c] --> c 
    if!fac6220low : [a * b * c] --> c 
    le : [b * b] --> a 
    low : [b * c] --> c 
    map : [b -> b * c] --> c 
    minus : [b * b] --> b 
    nil : [] --> c 
    quicksort : [c] --> c 
    quot : [b * b] --> b 
    s : [b] --> b 
    true : [] --> a 

  Rules:

    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    quot(0, s(x)) => 0 
    quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) 
    le(0, x) => true 
    le(s(x), 0) => false 
    le(s(x), s(y)) => le(x, y) 
    app(nil, x) => x 
    app(add(x, y), z) => add(x, app(y, z)) 
    low(x, nil) => nil 
    low(x, add(y, z)) => if!fac6220low(le(y, x), x, add(y, z)) 
    if!fac6220low(true, x, add(y, z)) => add(y, low(x, z)) 
    if!fac6220low(false, x, add(y, z)) => low(x, z) 
    high(x, nil) => nil 
    high(x, add(y, z)) => if!fac6220high(le(y, x), x, add(y, z)) 
    if!fac6220high(true, x, add(y, z)) => high(x, z) 
    if!fac6220high(false, x, add(y, z)) => add(y, high(x, z)) 
    quicksort(nil) => nil 
    quicksort(add(x, y)) => app(quicksort(low(x, y)), add(x, quicksort(high(x, y)))) 
    map(f, nil) => nil 
    map(f, add(x, y)) => add(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, add(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => add(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, 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:

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
  le(0, X) => true 
  le(s(X), 0) => false 
  le(s(X), s(Y)) => le(X, Y) 
  app(nil, X) => X 
  app(add(X, Y), Z) => add(X, app(Y, Z)) 
  low(X, nil) => nil 
  low(X, add(Y, Z)) => if!fac6220low(le(Y, X), X, add(Y, Z)) 
  if!fac6220low(true, X, add(Y, Z)) => add(Y, low(X, Z)) 
  if!fac6220low(false, X, add(Y, Z)) => low(X, Z) 
  high(X, nil) => nil 
  high(X, add(Y, Z)) => if!fac6220high(le(Y, X), X, add(Y, Z)) 
  if!fac6220high(true, X, add(Y, Z)) => high(X, Z) 
  if!fac6220high(false, X, add(Y, Z)) => add(Y, high(X, Z)) 
  quicksort(nil) => nil 
  quicksort(add(X, Y)) => app(quicksort(low(X, Y)), add(X, quicksort(high(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) Overlay + Local Confluence [EQUIVALENT]
 || (2) QTRS
 || (3) DependencyPairsProof [EQUIVALENT]
 || (4) QDP
 || (5) DependencyGraphProof [EQUIVALENT]
 || (6) AND
 ||     (7) QDP
 ||         (8) UsableRulesProof [EQUIVALENT]
 ||         (9) QDP
 ||         (10) QReductionProof [EQUIVALENT]
 ||         (11) QDP
 ||         (12) QDPSizeChangeProof [EQUIVALENT]
 ||         (13) YES
 ||     (14) QDP
 ||         (15) UsableRulesProof [EQUIVALENT]
 ||         (16) QDP
 ||         (17) QReductionProof [EQUIVALENT]
 ||         (18) QDP
 ||         (19) QDPSizeChangeProof [EQUIVALENT]
 ||         (20) YES
 ||     (21) QDP
 ||         (22) UsableRulesProof [EQUIVALENT]
 ||         (23) QDP
 ||         (24) QReductionProof [EQUIVALENT]
 ||         (25) QDP
 ||         (26) QDPSizeChangeProof [EQUIVALENT]
 ||         (27) YES
 ||     (28) QDP
 ||         (29) UsableRulesProof [EQUIVALENT]
 ||         (30) QDP
 ||         (31) QReductionProof [EQUIVALENT]
 ||         (32) QDP
 ||         (33) QDPSizeChangeProof [EQUIVALENT]
 ||         (34) YES
 ||     (35) QDP
 ||         (36) UsableRulesProof [EQUIVALENT]
 ||         (37) QDP
 ||         (38) QReductionProof [EQUIVALENT]
 ||         (39) QDP
 ||         (40) QDPOrderProof [EQUIVALENT]
 ||         (41) QDP
 ||         (42) UsableRulesProof [EQUIVALENT]
 ||         (43) QDP
 ||         (44) QReductionProof [EQUIVALENT]
 ||         (45) QDP
 ||         (46) QDPOrderProof [EQUIVALENT]
 ||         (47) QDP
 ||         (48) PisEmptyProof [EQUIVALENT]
 ||         (49) YES
 ||     (50) QDP
 ||         (51) UsableRulesProof [EQUIVALENT]
 ||         (52) QDP
 ||         (53) QReductionProof [EQUIVALENT]
 ||         (54) QDP
 ||         (55) QDPSizeChangeProof [EQUIVALENT]
 ||         (56) YES
 ||     (57) QDP
 ||         (58) UsableRulesProof [EQUIVALENT]
 ||         (59) QDP
 ||         (60) QReductionProof [EQUIVALENT]
 ||         (61) QDP
 ||         (62) QDPOrderProof [EQUIVALENT]
 ||         (63) QDP
 ||         (64) PisEmptyProof [EQUIVALENT]
 ||         (65) YES
 || 
 || 
 || ----------------------------------------
 || 
 || (0)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || Q is empty.
 || 
 || ----------------------------------------
 || 
 || (1) Overlay + Local Confluence (EQUIVALENT)
 || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
 || ----------------------------------------
 || 
 || (2)
 || Obligation:
 || Q restricted rewrite system:
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (3) DependencyPairsProof (EQUIVALENT)
 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
 || ----------------------------------------
 || 
 || (4)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
 ||    QUOT(s(%X), s(%Y)) -> MINUS(%X, %Y)
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 ||    APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 ||    LOW(%X, add(%Y, %Z)) -> IF!FAC6220LOW(le(%Y, %X), %X, add(%Y, %Z))
 ||    LOW(%X, add(%Y, %Z)) -> LE(%Y, %X)
 ||    IF!FAC6220LOW(true, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 ||    IF!FAC6220LOW(false, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 ||    HIGH(%X, add(%Y, %Z)) -> IF!FAC6220HIGH(le(%Y, %X), %X, add(%Y, %Z))
 ||    HIGH(%X, add(%Y, %Z)) -> LE(%Y, %X)
 ||    IF!FAC6220HIGH(true, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 ||    IF!FAC6220HIGH(false, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 ||    QUICKSORT(add(%X, %Y)) -> APP(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(low(%X, %Y))
 ||    QUICKSORT(add(%X, %Y)) -> LOW(%X, %Y)
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(high(%X, %Y))
 ||    QUICKSORT(add(%X, %Y)) -> HIGH(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (5) DependencyGraphProof (EQUIVALENT)
 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 6 less nodes.
 || ----------------------------------------
 || 
 || (6)
 || Complex Obligation (AND)
 || 
 || ----------------------------------------
 || 
 || (7)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (8) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (9)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (10) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (11)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (12) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *APP(add(%X, %Y), %Z) -> APP(%Y, %Z)
 || The graph contains the following edges 1 > 1, 2 >= 2
 || 
 || 
 || ----------------------------------------
 || 
 || (13)
 || YES
 || 
 || ----------------------------------------
 || 
 || (14)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (15) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (16)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (17) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (18)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (19) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *LE(s(%X), s(%Y)) -> LE(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (20)
 || YES
 || 
 || ----------------------------------------
 || 
 || (21)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    HIGH(%X, add(%Y, %Z)) -> IF!FAC6220HIGH(le(%Y, %X), %X, add(%Y, %Z))
 ||    IF!FAC6220HIGH(true, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 ||    IF!FAC6220HIGH(false, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (22) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (23)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    HIGH(%X, add(%Y, %Z)) -> IF!FAC6220HIGH(le(%Y, %X), %X, add(%Y, %Z))
 ||    IF!FAC6220HIGH(true, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 ||    IF!FAC6220HIGH(false, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (24) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (25)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    HIGH(%X, add(%Y, %Z)) -> IF!FAC6220HIGH(le(%Y, %X), %X, add(%Y, %Z))
 ||    IF!FAC6220HIGH(true, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 ||    IF!FAC6220HIGH(false, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (26) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *HIGH(%X, add(%Y, %Z)) -> IF!FAC6220HIGH(le(%Y, %X), %X, add(%Y, %Z))
 || The graph contains the following edges 1 >= 2, 2 >= 3
 || 
 || 
 || *IF!FAC6220HIGH(true, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 || The graph contains the following edges 2 >= 1, 3 > 2
 || 
 || 
 || *IF!FAC6220HIGH(false, %X, add(%Y, %Z)) -> HIGH(%X, %Z)
 || The graph contains the following edges 2 >= 1, 3 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (27)
 || YES
 || 
 || ----------------------------------------
 || 
 || (28)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LOW(%X, add(%Y, %Z)) -> IF!FAC6220LOW(le(%Y, %X), %X, add(%Y, %Z))
 ||    IF!FAC6220LOW(true, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 ||    IF!FAC6220LOW(false, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (29) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (30)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LOW(%X, add(%Y, %Z)) -> IF!FAC6220LOW(le(%Y, %X), %X, add(%Y, %Z))
 ||    IF!FAC6220LOW(true, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 ||    IF!FAC6220LOW(false, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (31) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (32)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    LOW(%X, add(%Y, %Z)) -> IF!FAC6220LOW(le(%Y, %X), %X, add(%Y, %Z))
 ||    IF!FAC6220LOW(true, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 ||    IF!FAC6220LOW(false, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (33) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *LOW(%X, add(%Y, %Z)) -> IF!FAC6220LOW(le(%Y, %X), %X, add(%Y, %Z))
 || The graph contains the following edges 1 >= 2, 2 >= 3
 || 
 || 
 || *IF!FAC6220LOW(true, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 || The graph contains the following edges 2 >= 1, 3 > 2
 || 
 || 
 || *IF!FAC6220LOW(false, %X, add(%Y, %Z)) -> LOW(%X, %Z)
 || The graph contains the following edges 2 >= 1, 3 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (34)
 || YES
 || 
 || ----------------------------------------
 || 
 || (35)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(high(%X, %Y))
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(low(%X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (36) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (37)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(high(%X, %Y))
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(low(%X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (38) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (39)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(high(%X, %Y))
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(low(%X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (40) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(high(%X, %Y))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(QUICKSORT(x_1)) = x_1
 ||    POL(add(x_1, x_2)) = 1 + x_2
 ||    POL(false) = 0
 ||    POL(high(x_1, x_2)) = x_2
 ||    POL(if!fac6220high(x_1, x_2, x_3)) = x_3
 ||    POL(if!fac6220low(x_1, x_2, x_3)) = 1 + x_3
 ||    POL(le(x_1, x_2)) = 0
 ||    POL(low(x_1, x_2)) = 1 + x_2
 ||    POL(nil) = 0
 ||    POL(s(x_1)) = 0
 ||    POL(true) = 0
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 || 
 || 
 || ----------------------------------------
 || 
 || (41)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(low(%X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (42) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (43)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(low(%X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (44) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 || 
 || 
 || ----------------------------------------
 || 
 || (45)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(low(%X, %Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (46) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    QUICKSORT(add(%X, %Y)) -> QUICKSORT(low(%X, %Y))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
 || 
 || POL( QUICKSORT_1(x_1) ) = max{0, 2x_1 - 1}
 || POL( low_2(x_1, x_2) ) = x_2
 || POL( nil ) = 1
 || POL( add_2(x_1, x_2) ) = 2x_2 + 2
 || POL( if!fac6220low_3(x_1, ..., x_3) ) = max{0, x_1 + x_3 - 2}
 || POL( le_2(x_1, x_2) ) = 2
 || POL( false ) = 0
 || POL( 0 ) = 2
 || POL( true ) = 2
 || POL( s_1(x_1) ) = 2x_1 + 2
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 || 
 || 
 || ----------------------------------------
 || 
 || (47)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 || 
 || The set Q consists of the following terms:
 || 
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (48) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (49)
 || YES
 || 
 || ----------------------------------------
 || 
 || (50)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (51) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (52)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || R is empty.
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (53) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (54)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || 
 || R is empty.
 || Q is empty.
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (55) QDPSizeChangeProof (EQUIVALENT)
 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
 || 
 || From the DPs we obtained the following set of size-change graphs:
 || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
 || The graph contains the following edges 1 > 1, 2 > 2
 || 
 || 
 || ----------------------------------------
 || 
 || (56)
 || YES
 || 
 || ----------------------------------------
 || 
 || (57)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 ||    quot(0, s(%X)) -> 0
 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
 ||    le(0, %X) -> true
 ||    le(s(%X), 0) -> false
 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
 ||    app(nil, %X) -> %X
 ||    app(add(%X, %Y), %Z) -> add(%X, app(%Y, %Z))
 ||    low(%X, nil) -> nil
 ||    low(%X, add(%Y, %Z)) -> if!fac6220low(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220low(true, %X, add(%Y, %Z)) -> add(%Y, low(%X, %Z))
 ||    if!fac6220low(false, %X, add(%Y, %Z)) -> low(%X, %Z)
 ||    high(%X, nil) -> nil
 ||    high(%X, add(%Y, %Z)) -> if!fac6220high(le(%Y, %X), %X, add(%Y, %Z))
 ||    if!fac6220high(true, %X, add(%Y, %Z)) -> high(%X, %Z)
 ||    if!fac6220high(false, %X, add(%Y, %Z)) -> add(%Y, high(%X, %Z))
 ||    quicksort(nil) -> nil
 ||    quicksort(add(%X, %Y)) -> app(quicksort(low(%X, %Y)), add(%X, quicksort(high(%X, %Y))))
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (58) UsableRulesProof (EQUIVALENT)
 || As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
 || ----------------------------------------
 || 
 || (59)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (60) QReductionProof (EQUIVALENT)
 || We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
 || 
 ||    quot(0, s(x0))
 ||    quot(s(x0), s(x1))
 ||    le(0, x0)
 ||    le(s(x0), 0)
 ||    le(s(x0), s(x1))
 ||    app(nil, x0)
 ||    app(add(x0, x1), x2)
 ||    low(x0, nil)
 ||    low(x0, add(x1, x2))
 ||    if!fac6220low(true, x0, add(x1, x2))
 ||    if!fac6220low(false, x0, add(x1, x2))
 ||    high(x0, nil)
 ||    high(x0, add(x1, x2))
 ||    if!fac6220high(true, x0, add(x1, x2))
 ||    if!fac6220high(false, x0, add(x1, x2))
 ||    quicksort(nil)
 ||    quicksort(add(x0, x1))
 || 
 || 
 || ----------------------------------------
 || 
 || (61)
 || Obligation:
 || Q DP problem:
 || The TRS P consists of the following rules:
 || 
 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
 || 
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (62) QDPOrderProof (EQUIVALENT)
 || We use the reduction pair processor [LPAR04,JAR06].
 || 
 || 
 || The following pairs can be oriented strictly and are deleted.
 || 
 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
 || The remaining pairs can at least be oriented weakly.
 || Used ordering:  Polynomial interpretation [POLO]:
 || 
 ||    POL(0) = 0
 ||    POL(QUOT(x_1, x_2)) = x_1
 ||    POL(minus(x_1, x_2)) = x_1
 ||    POL(s(x_1)) = 1 + x_1
 || 
 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 || 
 || 
 || ----------------------------------------
 || 
 || (63)
 || Obligation:
 || Q DP problem:
 || P is empty.
 || The TRS R consists of the following rules:
 || 
 ||    minus(%X, 0) -> %X
 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
 || 
 || The set Q consists of the following terms:
 || 
 ||    minus(x0, 0)
 ||    minus(s(x0), s(x1))
 || 
 || We have to consider all minimal (P,Q,R)-chains.
 || ----------------------------------------
 || 
 || (64) PisEmptyProof (EQUIVALENT)
 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
 || ----------------------------------------
 || 
 || (65)
 || 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] map#(F, add(X, Y)) =#> map#(F, Y)   
    1] filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y)   
    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    quot(0, s(X)) => 0 
    quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
    le(0, X) => true 
    le(s(X), 0) => false 
    le(s(X), s(Y)) => le(X, Y) 
    app(nil, X) => X 
    app(add(X, Y), Z) => add(X, app(Y, Z)) 
    low(X, nil) => nil 
    low(X, add(Y, Z)) => if!fac6220low(le(Y, X), X, add(Y, Z)) 
    if!fac6220low(true, X, add(Y, Z)) => add(Y, low(X, Z)) 
    if!fac6220low(false, X, add(Y, Z)) => low(X, Z) 
    high(X, nil) => nil 
    high(X, add(Y, Z)) => if!fac6220high(le(Y, X), X, add(Y, Z)) 
    if!fac6220high(true, X, add(Y, Z)) => high(X, Z) 
    if!fac6220high(false, X, add(Y, Z)) => add(Y, high(X, Z)) 
    quicksort(nil) => nil 
    quicksort(add(X, Y)) => app(quicksort(low(X, Y)), add(X, quicksort(high(X, Y)))) 
    map(F, nil) => nil 
    map(F, add(X, Y)) => add(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, add(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => add(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, 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 : 0 
    * 1 : 2, 3 
    * 2 : 1 
    * 3 : 1 

This graph has the following strongly connected components:

  P_1:

    map#(F, add(X, Y)) =#> map#(F, Y)   

  P_2:

    filter#(F, add(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(F, 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 apply the subterm criterion with the following projection function:

  nu(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, add(X, Y))) = add(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 

By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(F, add(X, Y))) = add(X, Y) |> Y = nu(map#(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.