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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))


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(s(x), s(y)) → +1(x, y)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
*1(s(x), s(y)) → +1(x, y)
*1(s(x), s(y)) → *1(x, y)
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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

SUM(cons(x, l)) → +1(x, sum(l))
+1(s(x), s(y)) → +1(x, y)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
*1(s(x), s(y)) → +1(x, y)
*1(s(x), s(y)) → *1(x, y)
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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

+1(s(x), s(y)) → +1(x, y)
SUM(cons(x, l)) → +1(x, sum(l))
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
*1(s(x), s(y)) → +1(x, y)
*1(s(x), s(y)) → *1(x, y)
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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 4 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ 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:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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(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)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ 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:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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