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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

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))
SUM(nil) → 01(#)
+1(1(x), 1(y)) → +1(x, y)
SUM(cons(x, l)) → SUM(l)
+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → 01(+(+(x, y), 1(#)))
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)
*1(1(x), y) → 01(*(x, y))
*1(0(x), y) → 01(*(x, y))
*1(1(x), y) → *1(x, y)
+1(0(x), 0(y)) → 01(+(x, y))
*1(0(x), y) → *1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
*1(1(x), y) → +1(0(*(x, y)), y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

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

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))
SUM(nil) → 01(#)
+1(1(x), 1(y)) → +1(x, y)
SUM(cons(x, l)) → SUM(l)
+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → 01(+(+(x, y), 1(#)))
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)
*1(1(x), y) → 01(*(x, y))
*1(0(x), y) → 01(*(x, y))
*1(1(x), y) → *1(x, y)
+1(0(x), 0(y)) → 01(+(x, y))
*1(0(x), y) → *1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
*1(1(x), y) → +1(0(*(x, y)), y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

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

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))
SUM(nil) → 01(#)
+1(1(x), 1(y)) → +1(x, y)
SUM(cons(x, l)) → SUM(l)
+1(1(x), 1(y)) → 01(+(+(x, y), 1(#)))
+1(0(x), 0(y)) → +1(x, y)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)
*1(1(x), y) → 01(*(x, y))
*1(1(x), y) → *1(x, y)
*1(0(x), y) → 01(*(x, y))
*1(0(x), y) → *1(x, y)
+1(0(x), 0(y)) → 01(+(x, y))
+1(1(x), 0(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
*1(1(x), y) → +1(0(*(x, y)), y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 8 less nodes.

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

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

+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

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

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(1(x), 1(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.

+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x2
0(x1)  =  x1
1(x1)  =  1(x1)
#  =  #

Lexicographic path order with status [19].
Quasi-Precedence:
11 > #

Status:
trivial


The following usable rules [14] were oriented: none



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

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

+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

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

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

+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

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

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

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

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

+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)

The TRS R consists of the following rules:

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

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

Lexicographic path order with status [19].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



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

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

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

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:

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

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(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)

Lexicographic path order with status [19].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



↳ 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:

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

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

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

*1(1(x), y) → *1(x, y)
*1(0(x), y) → *1(x, y)

The TRS R consists of the following rules:

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

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(0(x), y) → *1(x, y)
The remaining pairs can at least be oriented weakly.

*1(1(x), y) → *1(x, y)
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  x1
1(x1)  =  x1
0(x1)  =  0(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



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

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

*1(1(x), y) → *1(x, y)

The TRS R consists of the following rules:

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

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(1(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)  =  x1
1(x1)  =  1(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



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

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

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

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

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:

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

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)  =  x1
cons(x1, x2)  =  cons(x2)

Lexicographic path order with status [19].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [14] were oriented: none



↳ 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:

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

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.