Termination w.r.t. Q proof of TRS/various10.trs

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:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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:

SUM(++(xs, :(x, :(y, ys)))) → ++¹(xs, sum(:(x, :(y, ys))))
AVG(xs) → HD(sum(xs))
QUOT(s(x), s(y)) → -¹(x, y)
LENGTH(:(x, xs)) → LENGTH(xs)
-¹(s(x), s(y)) → -¹(x, y)
+¹(s(x), y) → +¹(x, y)
AVG(xs) → QUOT(hd(sum(xs)), length(xs))
AVG(xs) → SUM(xs)
++¹(:(x, xs), ys) → ++¹(xs, ys)
AVG(xs) → LENGTH(xs)
SUM(:(x, :(y, xs))) → +¹(x, y)
SUM(:(x, :(y, xs))) → SUM(:(+(x, y), xs))
QUOT(s(x), s(y)) → QUOT(-(x, y), s(y))
SUM(++(xs, :(x, :(y, ys)))) → SUM(:(x, :(y, ys)))
SUM(++(xs, :(x, :(y, ys)))) → SUM(++(xs, sum(:(x, :(y, ys)))))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

SUM(++(xs, :(x, :(y, ys)))) → ++¹(xs, sum(:(x, :(y, ys))))
AVG(xs) → HD(sum(xs))
QUOT(s(x), s(y)) → -¹(x, y)
LENGTH(:(x, xs)) → LENGTH(xs)
-¹(s(x), s(y)) → -¹(x, y)
+¹(s(x), y) → +¹(x, y)
AVG(xs) → QUOT(hd(sum(xs)), length(xs))
AVG(xs) → SUM(xs)
++¹(:(x, xs), ys) → ++¹(xs, ys)
AVG(xs) → LENGTH(xs)
SUM(:(x, :(y, xs))) → +¹(x, y)
SUM(:(x, :(y, xs))) → SUM(:(+(x, y), xs))
QUOT(s(x), s(y)) → QUOT(-(x, y), s(y))
SUM(++(xs, :(x, :(y, ys)))) → SUM(:(x, :(y, ys)))
SUM(++(xs, :(x, :(y, ys)))) → SUM(++(xs, sum(:(x, :(y, ys)))))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

AVG(xs) → HD(sum(xs))
SUM(++(xs, :(x, :(y, ys)))) → ++¹(xs, sum(:(x, :(y, ys))))
LENGTH(:(x, xs)) → LENGTH(xs)
QUOT(s(x), s(y)) → -¹(x, y)
-¹(s(x), s(y)) → -¹(x, y)
+¹(s(x), y) → +¹(x, y)
AVG(xs) → QUOT(hd(sum(xs)), length(xs))
++¹(:(x, xs), ys) → ++¹(xs, ys)
AVG(xs) → SUM(xs)
SUM(:(x, :(y, xs))) → +¹(x, y)
AVG(xs) → LENGTH(xs)
SUM(:(x, :(y, xs))) → SUM(:(+(x, y), xs))
QUOT(s(x), s(y)) → QUOT(-(x, y), s(y))
SUM(++(xs, :(x, :(y, ys)))) → SUM(:(x, :(y, ys)))
SUM(++(xs, :(x, :(y, ys)))) → SUM(++(xs, sum(:(x, :(y, ys)))))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 7 SCCs with 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

LENGTH(:(x, xs)) → LENGTH(xs)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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 oriented strictly and are deleted.

LENGTH(:(x, xs)) → LENGTH(xs)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LENGTH(x1) = LENGTH(x1)
:(x1, x2) = :(x2)

Recursive Path Order [2].
Precedence:

[LENGTH₁, :_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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 oriented strictly and are deleted.

-¹(s(x), s(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2) = -¹(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

s₁ > -^1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

QUOT(s(x), s(y)) → QUOT(-(x, y), s(y))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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 oriented strictly and are deleted.

QUOT(s(x), s(y)) → QUOT(-(x, y), s(y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2) = QUOT(x1, x2)
s(x1) = s(x1)
-(x1, x2) = x1
0 = 0

Recursive Path Order [2].
Precedence:

QUOT₂ > s₁
0 > s₁

The following usable rules [14] were oriented:

-(0, s(y)) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

++¹(:(x, xs), ys) → ++¹(xs, ys)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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 oriented strictly and are deleted.

++¹(:(x, xs), ys) → ++¹(xs, ys)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
++¹(x1, x2) = x1
:(x1, x2) = :(x1, x2)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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 oriented strictly and are deleted.

+¹(s(x), y) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = +¹(x1)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

s₁ > +^1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

SUM(:(x, :(y, xs))) → SUM(:(+(x, y), xs))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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 oriented strictly and are deleted.

SUM(:(x, :(y, xs))) → SUM(:(+(x, y), xs))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SUM(x1) = x1
:(x1, x2) = :(x2)
+(x1, x2) = x2
s(x1) = s(x1)
0 = 0

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

SUM(++(xs, :(x, :(y, ys)))) → SUM(++(xs, sum(:(x, :(y, ys)))))

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
++(nil, ys) → ys
++(:(x, xs), ys) → :(x, ++(xs, ys))
sum(:(x, nil)) → :(x, nil)
sum(:(x, :(y, xs))) → sum(:(+(x, y), xs))
sum(++(xs, :(x, :(y, ys)))) → sum(++(xs, sum(:(x, :(y, ys)))))
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(-(x, y), s(y)))
length(nil) → 0
length(:(x, xs)) → s(length(xs))
hd(:(x, xs)) → x
avg(xs) → quot(hd(sum(xs)), length(xs))

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