Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(plus, x)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(plus, x), y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(quot, app'2(app'2(minus, x), y))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(s, app'2(app'2(plus, x), y))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(cons, x), app'2(app'2(app, l), k))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k))))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(plus, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(plus, y), z)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(plus, y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(cons, app'2(app'2(plus, x), y)), l)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(cons, app'2(app'2(plus, x), y))
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app, l)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(plus, x)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(plus, x), y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(quot, app'2(app'2(minus, x), y))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(s, app'2(app'2(plus, x), y))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(cons, x), app'2(app'2(app, l), k))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k))))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(plus, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(plus, y), z)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(plus, y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(cons, app'2(app'2(plus, x), y)), l)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(cons, app'2(app'2(plus, x), y))
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app, l)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 17 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP'2(x1, x2)  =  x1
app'2(x1, x2)  =  app'1(x2)
app  =  app
cons  =  cons

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP'2(x1, x2)  =  APP'1(x1)
app'2(x1, x2)  =  app'2(x1, x2)
plus  =  plus
s  =  s

Lexicographic Path Order [19].
Precedence:
APP'1 > app'2
plus > app'2
s > app'2

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP'2(x1, x2)  =  x1
app'2(x1, x2)  =  app'2(x1, x2)
minus  =  minus
s  =  s
plus  =  plus
0  =  0

Lexicographic Path Order [19].
Precedence:
app'2 > s
minus > s
plus > s
0 > s

The following usable rules [14] were oriented:

app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))

The TRS R consists of the following rules:

app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.