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:

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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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:

+12(j1(x), j1(y)) -> +12(x, y)
*12(*2(x, y), z) -> *12(y, z)
OPP1(j1(x)) -> OPP1(x)
+12(11(x), j1(y)) -> +12(x, y)
+12(j1(x), 11(y)) -> +12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, z))
-12(x, y) -> +12(x, opp1(y))
+12(j1(x), 11(y)) -> 011(+2(x, y))
+12(11(x), j1(y)) -> 011(+2(x, y))
+12(01(x), 01(y)) -> +12(x, y)
-12(x, y) -> OPP1(y)
*12(j1(x), y) -> *12(x, y)
OPP1(01(x)) -> OPP1(x)
+12(+2(x, y), z) -> +12(y, z)
*12(11(x), y) -> 011(*2(x, y))
*12(11(x), y) -> *12(x, y)
+12(j1(x), j1(y)) -> +12(+2(x, y), j1(#))
+12(01(x), 01(y)) -> 011(+2(x, y))
*12(j1(x), y) -> 011(*2(x, y))
*12(11(x), y) -> +12(01(*2(x, y)), y)
*12(j1(x), y) -> -12(01(*2(x, y)), y)
+12(11(x), 11(y)) -> +12(x, y)
OPP1(01(x)) -> 011(opp1(x))
+12(j1(x), 01(y)) -> +12(x, y)
+12(01(x), j1(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(x, +2(y, z))
OPP1(11(x)) -> OPP1(x)
*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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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:

+12(j1(x), j1(y)) -> +12(x, y)
*12(*2(x, y), z) -> *12(y, z)
OPP1(j1(x)) -> OPP1(x)
+12(11(x), j1(y)) -> +12(x, y)
+12(j1(x), 11(y)) -> +12(x, y)
*12(*2(x, y), z) -> *12(x, *2(y, z))
-12(x, y) -> +12(x, opp1(y))
+12(j1(x), 11(y)) -> 011(+2(x, y))
+12(11(x), j1(y)) -> 011(+2(x, y))
+12(01(x), 01(y)) -> +12(x, y)
-12(x, y) -> OPP1(y)
*12(j1(x), y) -> *12(x, y)
OPP1(01(x)) -> OPP1(x)
+12(+2(x, y), z) -> +12(y, z)
*12(11(x), y) -> 011(*2(x, y))
*12(11(x), y) -> *12(x, y)
+12(j1(x), j1(y)) -> +12(+2(x, y), j1(#))
+12(01(x), 01(y)) -> 011(+2(x, y))
*12(j1(x), y) -> 011(*2(x, y))
*12(11(x), y) -> +12(01(*2(x, y)), y)
*12(j1(x), y) -> -12(01(*2(x, y)), y)
+12(11(x), 11(y)) -> +12(x, y)
OPP1(01(x)) -> 011(opp1(x))
+12(j1(x), 01(y)) -> +12(x, y)
+12(01(x), j1(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(x, +2(y, z))
OPP1(11(x)) -> OPP1(x)
*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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 11 less nodes.

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

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

OPP1(01(x)) -> OPP1(x)
OPP1(11(x)) -> OPP1(x)
OPP1(j1(x)) -> OPP1(x)

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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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.


OPP1(01(x)) -> OPP1(x)
OPP1(11(x)) -> OPP1(x)
OPP1(j1(x)) -> OPP1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(01(x1)) = 2 + 2·x1   
POL(11(x1)) = 2 + 2·x1   
POL(OPP1(x1)) = 2·x1   
POL(j1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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

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

+12(j1(x), j1(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(x, y)
+12(11(x), j1(y)) -> +12(x, y)
+12(j1(x), 11(y)) -> +12(x, y)
+12(01(x), j1(y)) -> +12(x, y)
+12(j1(x), 01(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(x, +2(y, z))
+12(01(x), 01(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(j1(x), j1(y)) -> +12(+2(x, y), j1(#))
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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.


+12(j1(x), j1(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(x, y)
+12(11(x), j1(y)) -> +12(x, y)
+12(j1(x), 11(y)) -> +12(x, y)
+12(01(x), j1(y)) -> +12(x, y)
+12(j1(x), 01(y)) -> +12(x, y)
+12(j1(x), j1(y)) -> +12(+2(x, y), j1(#))
+12(01(x), 11(y)) -> +12(x, y)
+12(11(x), 01(y)) -> +12(x, y)
+12(11(x), 11(y)) -> +12(+2(x, y), 11(#))
The remaining pairs can at least be oriented weakly.

+12(+2(x, y), z) -> +12(x, +2(y, z))
+12(01(x), 01(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
Used ordering: Polynomial interpretation [21]:

POL(#) = 0   
POL(+2(x1, x2)) = x1 + x2   
POL(+12(x1, x2)) = 2·x1 + 2·x2   
POL(01(x1)) = x1   
POL(11(x1)) = 1 + x1   
POL(j1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(#, x) -> x
+2(01(x), 11(y)) -> 11(+2(x, y))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(x, #) -> x
+2(01(x), 01(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(01(x), j1(y)) -> j1(+2(x, y))
+2(11(x), 01(y)) -> 11(+2(x, y))
01(#) -> #
+2(11(x), j1(y)) -> 01(+2(x, y))



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

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

+12(01(x), 01(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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.


+12(01(x), 01(y)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 2   
POL(+2(x1, x2)) = 1 + x1 + 2·x2   
POL(+12(x1, x2)) = x1   
POL(01(x1)) = 2 + 2·x1   
POL(11(x1)) = 1 + 2·x1   
POL(j1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ 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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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

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

*12(*2(x, y), z) -> *12(y, z)
*12(j1(x), y) -> *12(x, y)
*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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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.


*12(*2(x, y), z) -> *12(y, z)
*12(j1(x), y) -> *12(x, y)
*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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(#) = 2   
POL(*2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(*12(x1, x2)) = 2·x1   
POL(+2(x1, x2)) = 2 + x1 + 2·x2   
POL(-2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(01(x1)) = 2 + x1   
POL(11(x1)) = 2 + x1   
POL(j1(x1)) = 2 + x1   
POL(opp1(x1)) = x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(01(x), j1(y)) -> j1(+2(x, y))
+2(j1(x), 01(y)) -> j1(+2(x, y))
+2(11(x), 11(y)) -> j1(+2(+2(x, y), 11(#)))
+2(j1(x), j1(y)) -> 11(+2(+2(x, y), j1(#)))
+2(11(x), j1(y)) -> 01(+2(x, y))
+2(j1(x), 11(y)) -> 01(+2(x, y))
+2(+2(x, y), z) -> +2(x, +2(y, z))
opp1(#) -> #
opp1(01(x)) -> 01(opp1(x))
opp1(11(x)) -> j1(opp1(x))
opp1(j1(x)) -> 11(opp1(x))
-2(x, y) -> +2(x, opp1(y))
*2(#, x) -> #
*2(01(x), y) -> 01(*2(x, y))
*2(11(x), y) -> +2(01(*2(x, y)), y)
*2(j1(x), y) -> -2(01(*2(x, y)), y)
*2(*2(x, y), z) -> *2(x, *2(y, z))

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.