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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

Q is empty.

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

SUM1(nil) -> 011(#)
*12(*2(x, y), z) -> *12(y, z)
PROD1(app2(l1, l2)) -> PROD1(l2)
SUM1(app2(l1, l2)) -> +12(sum1(l1), sum1(l2))
*12(x, +2(y, z)) -> *12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, z))
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
+12(01(x), 01(y)) -> +12(x, y)
*12(x, +2(y, z)) -> +12(*2(x, y), *2(x, z))
+12(+2(x, y), z) -> +12(y, z)
PROD1(app2(l1, l2)) -> PROD1(l1)
*12(11(x), y) -> 011(*2(x, y))
*12(11(x), y) -> *12(x, y)
+12(01(x), 01(y)) -> 011(+2(x, y))
*12(11(x), y) -> +12(01(*2(x, y)), y)
APP2(cons2(x, l1), l2) -> APP2(l1, l2)
SUM1(cons2(x, l)) -> +12(x, sum1(l))
+12(11(x), 11(y)) -> +12(x, y)
SUM1(app2(l1, l2)) -> SUM1(l1)
SUM1(app2(l1, l2)) -> SUM1(l2)
+12(+2(x, y), z) -> +12(x, +2(y, z))
SUM1(cons2(x, l)) -> SUM1(l)
PROD1(cons2(x, l)) -> *12(x, prod1(l))
PROD1(app2(l1, l2)) -> *12(prod1(l1), prod1(l2))
PROD1(cons2(x, l)) -> PROD1(l)
*12(x, +2(y, z)) -> *12(x, z)
*12(01(x), y) -> 011(*2(x, y))
*12(01(x), y) -> *12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, 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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

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(nil) -> 011(#)
*12(*2(x, y), z) -> *12(y, z)
PROD1(app2(l1, l2)) -> PROD1(l2)
SUM1(app2(l1, l2)) -> +12(sum1(l1), sum1(l2))
*12(x, +2(y, z)) -> *12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, z))
+12(11(x), 11(y)) -> 011(+2(+2(x, y), 11(#)))
+12(01(x), 01(y)) -> +12(x, y)
*12(x, +2(y, z)) -> +12(*2(x, y), *2(x, z))
+12(+2(x, y), z) -> +12(y, z)
PROD1(app2(l1, l2)) -> PROD1(l1)
*12(11(x), y) -> 011(*2(x, y))
*12(11(x), y) -> *12(x, y)
+12(01(x), 01(y)) -> 011(+2(x, y))
*12(11(x), y) -> +12(01(*2(x, y)), y)
APP2(cons2(x, l1), l2) -> APP2(l1, l2)
SUM1(cons2(x, l)) -> +12(x, sum1(l))
+12(11(x), 11(y)) -> +12(x, y)
SUM1(app2(l1, l2)) -> SUM1(l1)
SUM1(app2(l1, l2)) -> SUM1(l2)
+12(+2(x, y), z) -> +12(x, +2(y, z))
SUM1(cons2(x, l)) -> SUM1(l)
PROD1(cons2(x, l)) -> *12(x, prod1(l))
PROD1(app2(l1, l2)) -> *12(prod1(l1), prod1(l2))
PROD1(cons2(x, l)) -> PROD1(l)
*12(x, +2(y, z)) -> *12(x, z)
*12(01(x), y) -> 011(*2(x, y))
*12(01(x), y) -> *12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(01(x), 11(y)) -> +12(x, 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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 11 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP2(cons2(x, l1), l2) -> APP2(l1, l2)

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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(cons2(x, l1), l2) -> APP2(l1, l2)
Used argument filtering: APP2(x1, x2)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ 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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ 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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

SUM1(app2(l1, l2)) -> SUM1(l1)
SUM1(app2(l1, l2)) -> SUM1(l2)
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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SUM1(app2(l1, l2)) -> SUM1(l1)
SUM1(app2(l1, l2)) -> SUM1(l2)
Used argument filtering: SUM1(x1)  =  x1
app2(x1, x2)  =  app2(x1, x2)
cons2(x1, x2)  =  x2
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
          ↳ 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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SUM1(cons2(x, l)) -> SUM1(l)
Used argument filtering: SUM1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

*12(*2(x, y), z) -> *12(y, z)
*12(x, +2(y, z)) -> *12(x, z)
*12(11(x), y) -> *12(x, y)
*12(x, +2(y, z)) -> *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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof

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

PROD1(app2(l1, l2)) -> PROD1(l2)
PROD1(cons2(x, l)) -> PROD1(l)
PROD1(app2(l1, l2)) -> PROD1(l1)

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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROD1(app2(l1, l2)) -> PROD1(l2)
PROD1(app2(l1, l2)) -> PROD1(l1)
Used argument filtering: PROD1(x1)  =  x1
app2(x1, x2)  =  app2(x1, x2)
cons2(x1, x2)  =  x2
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof

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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(l2))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROD1(cons2(x, l)) -> PROD1(l)
Used argument filtering: PROD1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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))
*2(x, +2(y, z)) -> +2(*2(x, y), *2(x, z))
app2(nil, l) -> l
app2(cons2(x, l1), l2) -> cons2(x, app2(l1, l2))
sum1(nil) -> 01(#)
sum1(cons2(x, l)) -> +2(x, sum1(l))
sum1(app2(l1, l2)) -> +2(sum1(l1), sum1(l2))
prod1(nil) -> 11(#)
prod1(cons2(x, l)) -> *2(x, prod1(l))
prod1(app2(l1, l2)) -> *2(prod1(l1), prod1(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.