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.

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 contains 6 SCCs with 17 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)
Used argument filtering: APP'2(x1, x2)  =  x1
app'2(x1, x2)  =  app'1(x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
Used argument filtering: APP'2(x1, x2)  =  x1
app'2(x1, x2)  =  app'1(x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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))
Used argument filtering: APP'2(x1, x2)  =  x2
app'2(x1, x2)  =  app'1(x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ 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(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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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))
Used argument filtering: APP'2(x1, x2)  =  x1
app'2(x1, x2)  =  app'2(x1, x2)
minus  =  minus
s  =  s
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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.