Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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:

SUM1(++2(xs, :2(x, :2(y, ys)))) -> ++12(xs, sum1(:2(x, :2(y, ys))))
AVG1(xs) -> HD1(sum1(xs))
QUOT2(s1(x), s1(y)) -> -12(x, y)
LENGTH1(:2(x, xs)) -> LENGTH1(xs)
-12(s1(x), s1(y)) -> -12(x, y)
+12(s1(x), y) -> +12(x, y)
AVG1(xs) -> QUOT2(hd1(sum1(xs)), length1(xs))
AVG1(xs) -> SUM1(xs)
++12(:2(x, xs), ys) -> ++12(xs, ys)
AVG1(xs) -> LENGTH1(xs)
SUM1(:2(x, :2(y, xs))) -> +12(x, y)
SUM1(:2(x, :2(y, xs))) -> SUM1(:2(+2(x, y), xs))
QUOT2(s1(x), s1(y)) -> QUOT2(-2(x, y), s1(y))
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(:2(x, :2(y, ys)))
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(++2(xs, sum1(:2(x, :2(y, ys)))))

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))

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:

SUM1(++2(xs, :2(x, :2(y, ys)))) -> ++12(xs, sum1(:2(x, :2(y, ys))))
AVG1(xs) -> HD1(sum1(xs))
QUOT2(s1(x), s1(y)) -> -12(x, y)
LENGTH1(:2(x, xs)) -> LENGTH1(xs)
-12(s1(x), s1(y)) -> -12(x, y)
+12(s1(x), y) -> +12(x, y)
AVG1(xs) -> QUOT2(hd1(sum1(xs)), length1(xs))
AVG1(xs) -> SUM1(xs)
++12(:2(x, xs), ys) -> ++12(xs, ys)
AVG1(xs) -> LENGTH1(xs)
SUM1(:2(x, :2(y, xs))) -> +12(x, y)
SUM1(:2(x, :2(y, xs))) -> SUM1(:2(+2(x, y), xs))
QUOT2(s1(x), s1(y)) -> QUOT2(-2(x, y), s1(y))
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(:2(x, :2(y, ys)))
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(++2(xs, sum1(:2(x, :2(y, ys)))))

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

LENGTH1(:2(x, xs)) -> LENGTH1(xs)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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.


LENGTH1(:2(x, xs)) -> LENGTH1(xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(:2(x1, x2)) = 1 + x2   
POL(LENGTH1(x1)) = x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

-12(s1(x), s1(y)) -> -12(x, y)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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.


-12(s1(x), s1(y)) -> -12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-12(x1, x2)) = 3·x1 + 3·x2   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

QUOT2(s1(x), s1(y)) -> QUOT2(-2(x, y), s1(y))

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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.


QUOT2(s1(x), s1(y)) -> QUOT2(-2(x, y), s1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-2(x1, x2)) = 1 + x1   
POL(0) = 3   
POL(QUOT2(x1, x2)) = 3·x1   
POL(s1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented:

-2(s1(x), s1(y)) -> -2(x, y)
-2(x, 0) -> x
-2(0, s1(y)) -> 0



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

++12(:2(x, xs), ys) -> ++12(xs, ys)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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.


++12(:2(x, xs), ys) -> ++12(xs, ys)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(++12(x1, x2)) = x1   
POL(:2(x1, x2)) = 1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

+12(s1(x), y) -> +12(x, y)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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.


+12(s1(x), y) -> +12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+12(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

SUM1(:2(x, :2(y, xs))) -> SUM1(:2(+2(x, y), xs))

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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.


SUM1(:2(x, :2(y, xs))) -> SUM1(:2(+2(x, y), xs))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 2 + 2·x2   
POL(0) = 0   
POL(:2(x1, x2)) = 3 + 3·x1 + 3·x2   
POL(SUM1(x1)) = 2·x1   
POL(s1(x1)) = 1   

The following usable rules [14] were oriented:

+2(s1(x), y) -> s1(+2(x, y))
+2(0, y) -> y



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(++2(xs, sum1(:2(x, :2(y, ys)))))

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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.


SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(++2(xs, sum1(:2(x, :2(y, ys)))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 0   
POL(++2(x1, x2)) = 2·x1 + x2   
POL(0) = 0   
POL(:2(x1, x2)) = 2 + x2   
POL(SUM1(x1)) = x1   
POL(nil) = 0   
POL(s1(x1)) = 0   
POL(sum1(x1)) = 3   

The following usable rules [14] were oriented:

++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
++2(nil, ys) -> ys
sum1(:2(x, nil)) -> :2(x, nil)



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

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

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(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.