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:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(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:

SUM1(cons2(x, l)) -> +12(x, sum1(l))
SUM1(nil) -> 011(#)
*12(*2(x, y), z) -> *12(y, z)
+12(11(x), 11(y)) -> +12(x, y)
SUM1(cons2(x, l)) -> SUM1(l)
*12(*2(x, y), z) -> *12(x, *2(y, z))
+12(+2(x, y), z) -> +12(x, +2(y, z))
+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
PROD1(cons2(x, l)) -> *12(x, prod1(l))
PROD1(cons2(x, l)) -> PROD1(l)
+12(+2(x, y), z) -> +12(y, z)
*12(11(x), y) -> 011(*2(x, y))
*12(11(x), y) -> *12(x, y)
*12(01(x), y) -> 011(*2(x, y))
+12(01(x), 01(y)) -> 011(+2(x, y))
*12(01(x), y) -> *12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
*12(11(x), y) -> +12(01(*2(x, y)), y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))

The TRS R consists of the following rules:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(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:

SUM1(cons2(x, l)) -> +12(x, sum1(l))
SUM1(nil) -> 011(#)
*12(*2(x, y), z) -> *12(y, z)
+12(11(x), 11(y)) -> +12(x, y)
SUM1(cons2(x, l)) -> SUM1(l)
*12(*2(x, y), z) -> *12(x, *2(y, z))
+12(+2(x, y), z) -> +12(x, +2(y, z))
+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
PROD1(cons2(x, l)) -> *12(x, prod1(l))
PROD1(cons2(x, l)) -> PROD1(l)
+12(+2(x, y), z) -> +12(y, z)
*12(11(x), y) -> 011(*2(x, y))
*12(11(x), y) -> *12(x, y)
*12(01(x), y) -> 011(*2(x, y))
+12(01(x), 01(y)) -> 011(+2(x, y))
*12(01(x), y) -> *12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
*12(11(x), y) -> +12(01(*2(x, y)), y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))

The TRS R consists of the following rules:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

+12(01(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(x, +2(y, z))
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))

The TRS R consists of the following rules:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))

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

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

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

SUM1(cons2(x, l)) -> SUM1(l)

The TRS R consists of the following rules:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(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 strictly oriented and are deleted.


SUM1(cons2(x, l)) -> SUM1(l)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SUM1(x1)  =  SUM1(x1)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
cons2 > SUM1


The following usable rules [14] 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:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(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

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

*12(*2(x, y), z) -> *12(y, z)
*12(11(x), y) -> *12(x, y)
*12(01(x), y) -> *12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, z))

The TRS R consists of the following rules:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))

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

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

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

PROD1(cons2(x, l)) -> PROD1(l)

The TRS R consists of the following rules:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(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 strictly oriented and are deleted.


PROD1(cons2(x, l)) -> PROD1(l)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PROD1(x1)  =  PROD1(x1)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
cons2 > PROD1


The following usable rules [14] 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:

01(#) -> #
+2(x, #) -> x
+2(#, x) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
+2(11(x), 11(y)) -> 01(+2(+2(x, y), 11(#)))
+2(+2(x, y), z) -> +2(x, +2(y, z))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(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.