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:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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(cons(x, l)) → +1(x, sum(l))
*1(*(x, y), z) → *1(y, z)
PROD(app(l1, l2)) → PROD(l2)
SUM(app(l1, l2)) → SUM(l1)
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
SUM(app(l1, l2)) → +1(sum(l1), sum(l2))
SUM(app(l1, l2)) → SUM(l2)
*1(s(x), s(y)) → *1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
*1(*(x, y), z) → *1(x, *(y, z))
SUM(cons(x, l)) → SUM(l)
+1(s(x), s(y)) → +1(x, y)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(app(l1, l2)) → *1(prod(l1), prod(l2))
PROD(cons(x, l)) → PROD(l)
+1(+(x, y), z) → +1(y, z)
PROD(app(l1, l2)) → PROD(l1)
*1(s(x), s(y)) → +1(x, y)
APP(cons(x, l1), l2) → APP(l1, l2)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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(cons(x, l)) → +1(x, sum(l))
*1(*(x, y), z) → *1(y, z)
PROD(app(l1, l2)) → PROD(l2)
SUM(app(l1, l2)) → SUM(l1)
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
SUM(app(l1, l2)) → +1(sum(l1), sum(l2))
SUM(app(l1, l2)) → SUM(l2)
*1(s(x), s(y)) → *1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
*1(*(x, y), z) → *1(x, *(y, z))
SUM(cons(x, l)) → SUM(l)
+1(s(x), s(y)) → +1(x, y)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(app(l1, l2)) → *1(prod(l1), prod(l2))
PROD(cons(x, l)) → PROD(l)
+1(+(x, y), z) → +1(y, z)
PROD(app(l1, l2)) → PROD(l1)
*1(s(x), s(y)) → +1(x, y)
APP(cons(x, l1), l2) → APP(l1, l2)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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:

SUM(cons(x, l)) → +1(x, sum(l))
*1(*(x, y), z) → *1(y, z)
SUM(app(l1, l2)) → SUM(l1)
PROD(app(l1, l2)) → PROD(l2)
SUM(app(l1, l2)) → +1(sum(l1), sum(l2))
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
SUM(app(l1, l2)) → SUM(l2)
*1(s(x), s(y)) → *1(x, y)
SUM(cons(x, l)) → SUM(l)
*1(*(x, y), z) → *1(x, *(y, z))
+1(+(x, y), z) → +1(x, +(y, z))
+1(s(x), s(y)) → +1(x, y)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)
PROD(app(l1, l2)) → *1(prod(l1), prod(l2))
+1(+(x, y), z) → +1(y, z)
PROD(app(l1, l2)) → PROD(l1)
*1(s(x), s(y)) → +1(x, y)
APP(cons(x, l1), l2) → APP(l1, l2)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 6 less nodes.

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

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

APP(cons(x, l1), l2) → APP(l1, l2)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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.


APP(cons(x, l1), l2) → APP(l1, l2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

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

+1(s(x), s(y)) → +1(x, y)
+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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)
+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
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)
+(x1, x2)  =  +(x1, x2)
0  =  0

Recursive Path Order [2].
Precedence:
s1 > [+^11, +2]
0 > [+^11, +2]


The following usable rules [14] were oriented: none



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

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

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

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

SUM(app(l1, l2)) → SUM(l1)
SUM(app(l1, l2)) → SUM(l2)
SUM(cons(x, l)) → SUM(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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(app(l1, l2)) → SUM(l1)
SUM(app(l1, l2)) → SUM(l2)
SUM(cons(x, l)) → SUM(l)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  SUM(x1)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [2].
Precedence:
[SUM1, app2]


The following usable rules [14] were oriented: none



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

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

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

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

*1(*(x, y), z) → *1(y, z)
*1(s(x), s(y)) → *1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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, y), z) → *1(y, z)
*1(s(x), s(y)) → *1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))
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)
s(x1)  =  s(x1)
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive Path Order [2].
Precedence:
[*2, s1, 0, +2]


The following usable rules [14] were oriented: none



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

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

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

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

PROD(app(l1, l2)) → PROD(l2)
PROD(cons(x, l)) → PROD(l)
PROD(app(l1, l2)) → PROD(l1)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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.


PROD(app(l1, l2)) → PROD(l2)
PROD(app(l1, l2)) → PROD(l1)
The remaining pairs can at least be oriented weakly.

PROD(cons(x, l)) → PROD(l)
Used ordering: Combined order from the following AFS and order.
PROD(x1)  =  PROD(x1)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  x2

Recursive Path Order [2].
Precedence:
[PROD1, app2]


The following usable rules [14] were oriented: none



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

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

PROD(cons(x, l)) → PROD(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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.


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

Recursive Path Order [2].
Precedence:
[PROD1, cons1]


The following usable rules [14] were oriented: none



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

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

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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.