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:

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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

+1(0(x), 0(y)) → 01(+(x, y))
SUM(nil) → 01(#)
PROD(cons(x, l)) → PROD(l)
*1(0(x), y) → 01(*(x, y))
SUM(cons(x, l)) → +1(x, sum(l))
+1(1(x), 1(y)) → +1(x, y)
+1(1(x), 1(y)) → 01(+(+(x, y), 1(#)))
+1(0(x), 0(y)) → +1(x, y)
*1(1(x), y) → +1(0(*(x, y)), y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
*1(1(x), y) → 01(*(x, y))
SUM(cons(x, l)) → SUM(l)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
PROD(cons(x, l)) → *1(x, prod(l))
*1(0(x), y) → *1(x, y)
*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.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 8 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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), 0(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(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 [15].


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.

+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = (4)x_1   
POL(0(x1)) = 1/4 + (4)x_1   
POL(+1(x1, x2)) = (1/4)x_2   
POL(+(x1, x2)) = 0   
POL(#) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



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

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

+1(0(x), 1(y)) → +1(x, y)
+1(1(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 [15].


The following pairs can be oriented strictly and are deleted.


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

+1(1(x), 1(y)) → +1(+(x, y), 1(#))
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 1/4 + (4)x_1   
POL(0(x1)) = 0   
POL(+1(x1, x2)) = (2)x_2   
POL(+(x1, x2)) = 0   
POL(#) = 0   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


+1(1(x), 1(y)) → +1(+(x, y), 1(#))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 2 + (2)x_1   
POL(0(x1)) = (2)x_1   
POL(+1(x1, x2)) = (1/4)x_1 + (1/2)x_2   
POL(+(x1, x2)) = x_1 + x_2   
POL(#) = 0   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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
      ↳ 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 [15].


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: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 1/4 + (4)x_2   
POL(SUM(x1)) = (4)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



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

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

*1(0(x), y) → *1(x, y)
*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 [15].


The following pairs can be oriented strictly and are deleted.


*1(0(x), y) → *1(x, y)
*1(1(x), y) → *1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 4 + (4)x_1   
POL(*1(x1, x2)) = (4)x_1   
POL(0(x1)) = 1 + x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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
      ↳ 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 [15].


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: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 2 + (4)x_2   
POL(PROD(x1)) = (4)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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.